There is a directed graph with N vertices and M edges. The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i. We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0. The player can move the piece as follows:

* When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i.



The player can end the game only when the piece is placed at vertex N. The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be? If it is possible to increase the score indefinitely, print `inf`.

Constraints

* 2≤N≤1000
* 1≤M≤min(N(N-1),2000)
* 1≤a_i,b_i≤N (1≤i≤M)
* a_i≠b_i (1≤i≤M)
* a_i≠a_j or b_i≠b_j (1≤i<j≤M)
* -10^9≤c_i≤10^9 (1≤i≤M)
* c_i is an integer.
* In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M


Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print `inf`.

Examples

Input

3 3
1 2 4
2 3 3
1 3 5


Output

7


Input

2 2
1 2 1
2 1 1


Output

inf


Input

6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000


Output

-5000000000

import java.io.*;
import java.util.*;

class Graph
{
    //開始,ゴール,頂点,辺
    int s, g, n, e;
    //無向の場合もコンストラクタでうまくやってくれるはず
    int[] f, t;
    long[] c;
    long d[];/*sからの最短距離*/
    long d2[][];/*2点間の最短距離*/

    public Graph(int s, int N, int[] f, int[] t, long[] c)
    {
        this.s = s;
        this.n = N;
        this.f = f;
        this.t = t;
        this.c = c;
        this.e = f.length;
    }

    //O( EV )type 0 : sから到達できなくても、負経路があればtrueを返す
    //       type 1 : sから到達できる　　　　負経路があればtrueを返す
    public boolean bellmanFord(int s, int type)
    {
        this.s = s;
        d = new long[n];
        if (type == 0)
        {
            Arrays.fill(d, 0);
        }
        else
        {
            Arrays.fill(d, Main.LINF);
            d[s] = 0;
        }
        //負経路がなければn-1回で最短が求まるため、N回目に更新されたら不経路がある
        for (int i = 0; i < n; i++)
        {
            for (int ei = 0; ei < e; ei++)
            {
                int f_ = f[ei];
                int t_ = t[ei];
                long c_ = c[ei];
                if (d[f_] != Main.LINF && d[t_] > d[f_] + c_)
                {
                    d[t_] = d[f_] + c_;
                    if (i == n - 1) return true;
                }
            }
        }
        return false;
    }

    //O( EV ) sからgの中に不経路があればtrueを返す
    public boolean bellmanFordSubset(int s, int g)
    {
        this.s = s;
        this.g = g;
        d = new long[n];
        Arrays.fill(d, Main.MINF);
        d[s] = 0;
        //負経路がなければn-1回で最短が求まるため、n～2N回目にgが更新されたら、s-gに不経路がある
        for (int i = 0; i < 2 * n; i++)
        {
            for (int ei = 0; ei < e; ei++)
            {
                int f_ = f[ei];
                int t_ = t[ei];
                long c_ = c[ei];
                if (d[f_] != Main.MINF && d[t_] < d[f_] + c_)
                {
                    d[t_] = d[f_] + c_;
                    if (i >= n - 1 && t_ == g) return true;
                }
            }
        }
        return false;
    }

    //O( E log V)
    public void dijkstra(int s)
    {
        class Edge implements Comparable<Edge>
        {
            int to;
            long cos;

            Edge(int to, long cost)
            {
                this.to = to;
                this.cos = cost;
            }

            public int compareTo(Edge ed)
            {
                if (cos != ed.cos)
                {
                    if (cos > ed.cos) return 1;
                    else if (cos == ed.cos) return 0;
                    else return -1;
                }
                else return to - ed.to;
            }
        }
        this.s = s;
        d = new long[n];
        List<ArrayList<Edge>> elist = new ArrayList<>();
        for (int i = 0; i < n; i++)
            elist.add(new ArrayList<>());

        Arrays.fill(d, Main.LINF);
        d[0] = 0;
        for (int ei = 0; ei < e; ei++)
        {
            int f_ = f[ei];
            int t_ = t[ei];
            long c_ = c[ei];
            elist.get(f_).add(new Edge(t_, c_));
        }
        PriorityQueue<Edge> q = new PriorityQueue<>();
        q.add(new Edge(s, 0));
        while (!q.isEmpty())
        {
            Edge p = q.poll();
            int nowt = p.to;
            long nowc = p.cos;
            if (d[nowt] < nowc) continue;
            for (int ti = 0; ti < elist.get(nowt).size(); ti++)
            {
                Edge ne = elist.get(nowt).get(ti);
                int newt = ne.to;
                long newc = nowc + ne.cos;
                if (newc < d[newt])
                {
                    d[newt] = newc;
                    q.add(new Edge(newt, newc));
                }
            }
        }

    }

    //O( E ^ 2) 辺が密の時用
    public void dijkstra2(int s)
    {
        this.s = s;
        boolean[] used = new boolean[e];
        long[][] cost = new long[n][n];
        d = new long[n];
        Main.fill(cost, Main.LINF);
        Arrays.fill(d, Main.LINF);
        d[s] = 0;

        for (int i = 0; i < e; i++)
            cost[f[i]][t[i]] = c[i];

        while (true)
        {
            int v = -1;
            //まだ使われていない頂点のうち、距離が最小のものを探す。
            for (int u = 0; u < n; u++)
                if (!used[u] && (v == -1 || d[u] < d[v])) v = u;
            if (v == -1) break;
            used[v] = true;
            for (int u = 0; u < n; u++)
                d[u] = Math.min(d[u], d[v] + cost[v][u]);
        }

    }

    public void warshall()
    {
        d2 = new long[n][n];
        Main.fill(d2, Main.LINF);
        for (int i = 0; i < n; i++)
            d2[i][i] = 0;
        for (int ei = 0; ei < e; ei++)
            d2[f[ei]][t[ei]] = Math.min(d2[f[ei]][t[ei]], c[ei]);
        for (int k = 0; k < n; k++)
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    d2[i][j] = Math.min(d2[i][j], d2[i][k] + d2[k][j]);
    }

    public long dis(int to)
    {
        return d[to];
    }

    public long dis2(int from, int to)
    {
        return d2[from][to];
    }
}

/**
 * @author baito
 */
public class Main
{
    static StringBuilder sb = new StringBuilder();
    static FastScanner sc = new FastScanner(System.in);
    static int INF = 12345678;
    static long LINF = 123456789123456789L;
    static long MINF = -123456789123456789L;
    static long MOD = 1000000007;
    static int[] y4 = {0, 1, 0, -1};
    static int[] x4 = {1, 0, -1, 0};
    static int[] y8 = {0, 1, 0, -1, -1, 1, 1, -1};
    static int[] x8 = {1, 0, -1, 0, 1, -1, 1, -1};
    static long[] F;//factorial
    static boolean[] isPrime;
    static int[] primes;

    static int N, M;


    public static void main(String[] args)
    {
        N = sc.nextInt();
        M = sc.nextInt();
        int[] a = new int[M];
        int[] b = new int[M];
        long[] c = new long[M];
        for (int i = 0; i < M; i++)
        {
            a[i] = sc.nextInt()-1;
            b[i] = sc.nextInt()-1;
            c[i] = sc.nextLong();
        }

        Graph g = new Graph(0, N, a, b, c);
        if (g.bellmanFordSubset(0, N - 1)) System.out.println("inf");
        else System.out.println(g.dis(N - 1));


    }

    public static void revSort(int[] a)
    {
        Arrays.sort(a);
        reverse(a);
    }

    public static void revSort(long[] a)
    {
        Arrays.sort(a);
        reverse(a);
    }

    public static int[][] copy(int[][] ar)
    {
        int[][] nr = new int[ar.length][ar[0].length];
        for (int i = 0; i < ar.length; i++)
            for (int j = 0; j < ar[0].length; j++)
                nr[i][j] = ar[i][j];
        return nr;
    }

    public static long sumMod(long... lar)
    {
        long sum = 0;
        for (long l : lar)
            sum = (sum + l % MOD) % MOD;
        return sum;
    }

    /**
     * <h1>指定した値以上の先頭のインデクスを返す</h1>
     * <p>配列要素が０のときは、０が返る。</p>
     *
     * @return<b>int</b> ： 探索した値以上で、先頭になるインデクス
     */
    public static int lowerBound(final int[] arr, final int value)
    {
        int low = 0;
        int high = arr.length;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >>> 1) + low;    //(low + high) / 2 (オーバーフロー対策)
            if (arr[mid] < value)
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
        }
        return low;
    }

    /**
     * <h1>指定した値より大きい先頭のインデクスを返す</h1>
     * <p>配列要素が０のときは、０が返る。</p>
     *
     * @return<b>int</b> ： 探索した値より上で、先頭になるインデクス
     */
    public static int upperBound(final int[] arr, final int value)
    {
        int low = 0;
        int high = arr.length;
        int mid;
        while (low < high)
        {
            mid = ((high - low) >>> 1) + low;    //(low + high) / 2 (オーバーフロー対策)
            if (arr[mid] <= value)
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
        }
        return low;
    }

    //次の順列に書き換える、最大値ならfalseを返す
    public static boolean nextPermutation(int A[])
    {
        int len = A.length;
        int pos = len - 2;
        for (; pos >= 0; pos--)
        {
            if (A[pos] < A[pos + 1]) break;
        }
        if (pos == -1) return false;

        //posより大きい最小の数を二分探索
        int ok = pos + 1;
        int ng = len;
        while (Math.abs(ng - ok) > 1)
        {
            int mid = (ok + ng) / 2;
            if (A[mid] > A[pos]) ok = mid;
            else ng = mid;

        }

        swap(A, pos, ok);
        reverse(A, pos + 1, len - 1);


        return true;
    }

    //次の順列に書き換える、最小値ならfalseを返す
    public static boolean prevPermutation(int A[])
    {
        int len = A.length;
        int pos = len - 2;
        for (; pos >= 0; pos--)
        {
            if (A[pos] > A[pos + 1]) break;
        }
        if (pos == -1) return false;

        //posより小さい最大の数を二分探索
        int ok = pos + 1;
        int ng = len;
        while (Math.abs(ng - ok) > 1)
        {
            int mid = (ok + ng) / 2;
            if (A[mid] < A[pos]) ok = mid;
            else ng = mid;

        }

        swap(A, pos, ok);
        reverse(A, pos + 1, len - 1);


        return true;
    }

    //↓nCrをmod計算するために必要。　***factorial(N)を呼ぶ必要がある***
    static long ncr(int n, int r)
    {
        if (n < r) return 0;
        else if (r == 0) return 1;

        factorial(n);
        return F[n] / (F[n - r] * F[r]);
    }

    static long ncr2(int a, int b)
    {
        if (b == 0) return 1;
        else if (a < b) return 0;
        long res = 1;
        for (int i = 0; i < b; i++)
        {
            res *= a - i;
            res /= i + 1;
        }
        return res;
    }

    static long ncrdp(int n, int r)
    {
        if (n < r) return 0;
        long[][] dp = new long[n + 1][r + 1];
        for (int ni = 0; ni < n + 1; ni++)
        {
            dp[ni][0] = dp[ni][ni] = 1;
            for (int ri = 1; ri < ni; ri++)
            {
                dp[ni][ri] = dp[ni - 1][ri - 1] + dp[ni - 1][ri];
            }
        }
        return dp[n][r];
    }

    static long modNcr(int n, int r)
    {
        long result = F[n];
        result = result * modInv(F[n - r]) % MOD;
        result = result * modInv(F[r]) % MOD;
        return result;
    }

    static long modInv(long n)
    {
        return modPow(n, MOD - 2);
    }

    static void factorial(int n)
    {
        F = new long[n + 1];
        F[0] = F[1] = 1;
        for (int i = 2; i <= n; i++)
        {
            F[i] = (F[i - 1] * i) % MOD;
        }
    }

    static long modPow(long x, long n)
    {
        long res = 1L;
        while (n > 0)
        {
            if ((n & 1) == 1)
            {
                res = res * x % MOD;
            }
            x = x * x % MOD;
            n >>= 1;
        }
        return res;
    }

    //↑nCrをmod計算するために必要

    static int gcd(int n, int r)
    {
        return r == 0 ? n : gcd(r, n % r);
    }

    static long gcd(long n, long r)
    {
        return r == 0 ? n : gcd(r, n % r);
    }

    static <T> void swap(T[] x, int i, int j)
    {
        T t = x[i];
        x[i] = x[j];
        x[j] = t;
    }

    static void swap(int[] x, int i, int j)
    {
        int t = x[i];
        x[i] = x[j];
        x[j] = t;
    }

    public static void reverse(int[] x)
    {
        int l = 0;
        int r = x.length - 1;
        while (l < r)
        {
            int temp = x[l];
            x[l] = x[r];
            x[r] = temp;
            l++;
            r--;
        }
    }

    public static void reverse(long[] x)
    {
        int l = 0;
        int r = x.length - 1;
        while (l < r)
        {
            long temp = x[l];
            x[l] = x[r];
            x[r] = temp;
            l++;
            r--;
        }
    }

    public static void reverse(int[] x, int s, int e)
    {
        int l = s;
        int r = e;
        while (l < r)
        {
            int temp = x[l];
            x[l] = x[r];
            x[r] = temp;
            l++;
            r--;
        }
    }

    static int length(int a)
    {
        int cou = 0;
        while (a != 0)
        {
            a /= 10;
            cou++;
        }
        return cou;
    }

    static int length(long a)
    {
        int cou = 0;
        while (a != 0)
        {
            a /= 10;
            cou++;
        }
        return cou;
    }

    static int countC2(char[][] a, char c)
    {
        int co = 0;
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                if (a[i][j] == c) co++;
        return co;
    }

    static int countI(int[] a, int key)
    {
        int co = 0;
        for (int i = 0; i < a.length; i++)
            if (a[i] == key) co++;
        return co;
    }

    static int countI(int[][] a, int key)
    {
        int co = 0;
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                if (a[i][j] == key) co++;
        return co;
    }

    static void fill(int[][] a, int v)
    {
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                a[i][j] = v;
    }

    static void fill(long[][] a, long v)
    {
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                a[i][j] = v;
    }

    static void fill(int[][][] a, int v)
    {
        for (int i = 0; i < a.length; i++)
            for (int j = 0; j < a[0].length; j++)
                for (int k = 0; k < a[0][0].length; k++)
                    a[i][j][k] = v;
    }

    static int max(int a, int b, int c)
    {
        return Math.max(a, Math.max(b, c));
    }

    static int max(int[] ar)
    {
        int res = Integer.MIN_VALUE;
        for (int i : ar)
            res = Math.max(res, i);
        return res;
    }

    static int max(int[][] ar)
    {
        int res = Integer.MIN_VALUE;
        for (int i[] : ar)
            res = max(i);
        return res;
    }

    static int min(int a, int b, int c)
    {
        return Math.min(a, Math.min(b, c));
    }

    static int min(int[] ar)
    {
        int res = Integer.MAX_VALUE;
        for (int i : ar)
            res = Math.min(res, i);
        return res;
    }

    static int min(int[][] ar)
    {
        int res = Integer.MAX_VALUE;
        for (int i[] : ar)
            res = min(i);
        return res;
    }

    static int abs(int a)
    {
        return Math.abs(a);
    }

    static class FastScanner
    {

        private BufferedReader reader = null;
        private StringTokenizer tokenizer = null;

        public FastScanner(InputStream in)
        {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        public String next()
        {
            if (tokenizer == null || !tokenizer.hasMoreTokens())
            {
                try
                {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e)
                {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        /*public String nextChar(){
            return (char)next()[0];
        }*/
        public String nextLine()
        {
            if (tokenizer == null || !tokenizer.hasMoreTokens())
            {
                try
                {
                    return reader.readLine();
                } catch (IOException e)
                {
                    throw new RuntimeException(e);
                }
            }

            return tokenizer.nextToken("\n");
        }

        public long nextLong()
        {
            return Long.parseLong(next());
        }

        public int nextInt()
        {
            return Integer.parseInt(next());
        }

        public double nextDouble()
        {
            return Double.parseDouble(next());
        }

        public int[] nextIntArray(int n)
        {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextInt();
            }
            return a;
        }

        public int[][] nextIntArray2(int h, int w)
        {
            int[][] a = new int[h][w];
            for (int hi = 0; hi < h; hi++)
            {
                for (int wi = 0; wi < w; wi++)
                {
                    a[hi][wi] = nextInt();
                }
            }
            return a;
        }

        //縦２つの列を一つにまとめる
        public int[] nextIntArray21(int n, int scalar)
        {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt() * scalar + nextInt();
            return a;
        }

        //複数の配列を受け取る
        public void nextIntArrays2ar(int n, int[] a, int[] b)
        {
            for (int i = 0; i < n; i++)
            {
                a[i] = sc.nextInt();
                b[i] = sc.nextInt();
            }
        }

        //複数の配列を受け取る
        public void nextIntArrays3ar(int n, int[] a, int[] b, int[] c)
        {
            for (int i = 0; i < n; i++)
            {
                a[i] = sc.nextInt();
                b[i] = sc.nextInt();
                c[i] = sc.nextInt();
            }
        }

        public Integer[] nextIntegerArray(int n)
        {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextInt();
            }
            return a;
        }

        public char[] nextCharArray(int n)
        {
            char[] a = next().toCharArray();

            return a;
        }

        public char[][] nextCharArray2(int h, int w)
        {
            char[][] a = new char[h][w];
            for (int i = 0; i < h; i++)
            {
                a[i] = next().toCharArray();
            }
            return a;
        }

        //スペースが入っている場合
        public char[][] nextCharArray2s(int h, int w)
        {
            char[][] a = new char[h][w];
            for (int i = 0; i < h; i++)
            {
                a[i] = nextLine().replace(" ", "").toCharArray();
            }
            return a;
        }

        public char[][] nextWrapCharArray2(int h, int w, char c)
        {
            char[][] a = new char[h + 2][w + 2];
            //char c = '*';
            int i;
            for (i = 0; i < w + 2; i++)
                a[0][i] = c;
            for (i = 1; i < h + 1; i++)
            {
                a[i] = (c + next() + c).toCharArray();
            }
            for (i = 0; i < w + 2; i++)
                a[h + 1][i] = c;
            return a;
        }

        //スペースが入ってる時用
        public char[][] nextWrapCharArray2s(int h, int w, char c)
        {
            char[][] a = new char[h + 2][w + 2];
            //char c = '*';
            int i;
            for (i = 0; i < w + 2; i++)
                a[0][i] = c;
            for (i = 1; i < h + 1; i++)
            {
                a[i] = (c + nextLine().replace(" ", "") + c).toCharArray();
            }
            for (i = 0; i < w + 2; i++)
                a[h + 1][i] = c;
            return a;
        }

        public long[] nextLongArray(int n)
        {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
            {
                a[i] = nextLong();
            }
            return a;
        }

        public long[][] nextLongArray2(int h, int w)
        {
            long[][] a = new long[h][w];
            for (int hi = 0; hi < h; hi++)
            {
                for (int wi = 0; wi < h; wi++)
                {
                    a[h][w] = nextLong();
                }
            }
            return a;
        }
    }
}
