Each evening after the dinner the SIS's students gather together to play the game of Sport Mafia. 

For the tournament, Alya puts candies into the box, which will serve as a prize for a winner. To do that, she performs n actions. The first action performed is to put a single candy into the box. For each of the remaining moves she can choose from two options:

  * the first option, in case the box contains at least one candy, is to take exactly one candy out and eat it. This way the number of candies in the box decreased by 1; 
  * the second option is to put candies in the box. In this case, Alya will put 1 more candy, than she put in the previous time. 



Thus, if the box is empty, then it can only use the second option.

For example, one possible sequence of Alya's actions look as follows:

  * put one candy into the box; 
  * put two candies into the box; 
  * eat one candy from the box; 
  * eat one candy from the box; 
  * put three candies into the box; 
  * eat one candy from the box; 
  * put four candies into the box; 
  * eat one candy from the box; 
  * put five candies into the box; 



This way she will perform 9 actions, the number of candies at the end will be 11, while Alya will eat 4 candies in total.

You know the total number of actions n and the number of candies at the end k. You need to find the total number of sweets Alya ate. That is the number of moves of the first option. It's guaranteed, that for the given n and k the answer always exists.

Please note, that during an action of the first option, Alya takes out and eats exactly one candy.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10^9; 0 ≤ k ≤ 10^9) — the total number of moves and the number of candies in the box at the end. 

It's guaranteed, that for the given n and k the answer exists.

Output

Print a single integer — the number of candies, which Alya ate. Please note, that in this problem there aren't multiple possible answers — the answer is unique for any input data. 

Examples

Input


1 1


Output


0

Input


9 11


Output


4

Input


5 0


Output


3

Input


3 2


Output


1

Note

In the first example, Alya has made one move only. According to the statement, the first move is always putting one candy in the box. Hence Alya ate 0 candies.

In the second example the possible sequence of Alya's actions looks as follows: 

  * put 1 candy, 
  * put 2 candies, 
  * eat a candy, 
  * eat a candy, 
  * put 3 candies, 
  * eat a candy, 
  * put 4 candies, 
  * eat a candy, 
  * put 5 candies. 



This way, she will make exactly n=9 actions and in the end the box will contain 1+2-1-1+3-1+4-1+5=11 candies. The answer is 4, since she ate 4 candies in total.
import java.io.*;
import java.util.*;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static final double EPS = 1e-9;
	static long mod = 1000000007;
	static int inf = (int) 1e9 + 2;
	static long[] fac;
	static int[] si;
	static ArrayList<Integer> primes;
	static ArrayList<Integer>[] ad;
	static boolean f;
    static long [][]memo;
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		long n=sc.nextInt();
		long k=sc.nextInt();
		long ans=0;
		long cons=-2*n-2*k;
		ans=(long)Math.max((-3+Math.sqrt(9-4*cons))/2,(-3-Math.sqrt(9-4*cons))/2);
		out.println(n-ans);
		out.close();
	}
   static long dp(int num,int k) {
	   if(num==0 || k==0 || k>num)
		   return 0;
	   if(num==k || k==1)
		   return 1;
	   if(memo[num][k]!=-1)
		   return memo[num][k];
	   long ans=((k*dp(num-1,k))%mod+dp(num-1,k-1))%mod;
	   return memo[num][k]=ans;
   }
	static class pair implements Comparable<pair> {
		int left, right, type;

		public pair(int x, int y, int t) {
			left = x;
			right = y;
			type = t;
		}

		public int compareTo(pair o) {
			if (type == 0) {
				if (o.left == left)
					return right - o.right;
				return left - o.left;
			} else {
				if (o.right == right)
					return o.left - left;
				return o.right - right;
			}
		}

		public String toString() {
			return left + " " + right;
		}
	}

	static class seg implements Comparable<seg> {
		int a;
		int b;

		seg(int s, int e) {
			a = s;
			b = e;
		}

		public String toString() {
			return a + " " + b;
		}

		public int compareTo(seg o) {
			// if(a==o.a)
			return o.b - b;
			// return
		}
	}

	static public class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy, candies;

		SegmentTree(int[] in) {
			array = in.clone();
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			candies = new int[N << 1];
			build(1, 1, N);
			build2(1, 1, N);
		}

		public String toString() {
			return Arrays.toString(sTree) + "\n" + Arrays.toString(candies);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = (sTree[node << 1] + sTree[node << 1 | 1]) % 10;
				if (sTree[node << 1] + sTree[node << 1 | 1] >= 10)
					candies[node]++;
			}
		}

		void build2(int node, int b, int e) // O(n)
		{
			if (b == e)
				b = e;
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				candies[node] = (candies[node << 1] + candies[node << 1 | 1]);
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while (index > 1) {
				index >>= 1;
				sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		// static int c=0;
		int query(int i, int j) {
			// c=0;
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return candies[node];
			int mid = b + e >> 1;
			// propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return q1 + q2;
		}
	}

	static int x, y, d;

	static long power(int i) {
		// if(i==0)
		// return 1;
		long a = 1;
		for (int k = 0; k < i; k++)
			a *= i;
		return a;
	}

	static void extendedEuclid(int a, int b) {
		if (b == 0) {
			x = 1;
			y = 0;
			d = a;
			return;
		}
		extendedEuclid(b, a % b);
		int x1 = y;
		int y1 = x - a / b * y;
		x = x1;
		y = y1;
	}

	static void seive() {
		si = new int[1000001];
		primes = new ArrayList<>();
		int N = 1000001;
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static long inver(long x) {
		int a = (int) x;
		long e = (mod - 2);
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				// System.out.println(res*a);
				res = (int) ((1l * res * a) % mod);
			}
			a = (int) ((1l * a * a) % mod);
			e >>= 1;
		}
		// out.println(res+" "+x);
		return res % mod;
	}

	static long fac(int n) {
		if (n == 0)
			return fac[n] = 1;
		if (n == 1)
			return fac[n] = 1;
		long ans = 1;
		for (int i = 1; i <= n; i++)
			fac[i] = ans = (i % mod * ans % mod) % mod;
		return ans % mod;
	}

	static int gcd(int a, int b) {

		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static class unionfind {
		int[] p;
		int[] size;

		unionfind(int n) {
			p = new int[n];
			size = new int[n];

			for (int i = 0; i < n; i++) {
				p[i] = i;
			}
			Arrays.fill(size, 1);
		}

		int findSet(int v) {
			if (v == p[v])
				return v;
			return p[v] = findSet(p[v]);
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		void combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return;
			if (size[a] > size[b]) {
				p[b] = a;
				size[a] += size[b];

			} else {
				p[a] = b;
				size[b] += size[a];
			}
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}