This morning, Roman woke up and opened the browser with n opened tabs numbered from 1 to n. There are two kinds of tabs: those with the information required for the test and those with social network sites. Roman decided that there are too many tabs open so he wants to close some of them.

He decided to accomplish this by closing every k-th (2 ≤ k ≤ n - 1) tab. Only then he will decide whether he wants to study for the test or to chat on the social networks. Formally, Roman will choose one tab (let its number be b) and then close all tabs with numbers c = b + i ⋅ k that satisfy the following condition: 1 ≤ c ≤ n and i is an integer (it may be positive, negative or zero).

For example, if k = 3, n = 14 and Roman chooses b = 8, then he will close tabs with numbers 2, 5, 8, 11 and 14.

After closing the tabs Roman will calculate the amount of remaining tabs with the information for the test (let's denote it e) and the amount of remaining social network tabs (s). Help Roman to calculate the maximal absolute value of the difference of those values |e - s| so that it would be easy to decide what to do next.

Input

The first line contains two integers n and k (2 ≤ k < n ≤ 100) — the amount of tabs opened currently and the distance between the tabs closed.

The second line consists of n integers, each of them equal either to 1 or to -1. The i-th integer denotes the type of the i-th tab: if it is equal to 1, this tab contains information for the test, and if it is equal to -1, it's a social network tab.

Output

Output a single integer — the maximum absolute difference between the amounts of remaining tabs of different types |e - s|.

Examples

Input


4 2
1 1 -1 1


Output


2


Input


14 3
-1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1


Output


9

Note

In the first example we can choose b = 1 or b = 3. We will delete then one tab of each type and the remaining tabs are then all contain test information. Thus, e = 2 and s = 0 and |e - s| = 2.

In the second example, on the contrary, we can leave opened only tabs that have social networks opened in them.


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

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 * @see format
 */
public class A {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static final int MOD = 1000000007;

    static int n, m, T;
    static long num;
    static int[] a, b, c;
    static long[] al, bl, cl;
    static String s, t;

    private static void solve() throws IOException {
        int max = 0;

        n = nextInt();
        int k = nextInt();
        a = new int[n];
        int sum = 0;
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            sum += a[i];
        }
        for (int i = 0; i < k; i++) {
            int cur = 0;
            for (int j = i; j < n; j += k) {
                cur += a[j];
            }
            max = Math.max(max, Math.abs(sum - cur));
        }

        pw.println(max);
    }

    static void swap(long a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void swap(int a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    // 18位素数：154590409516822759
    // 19位素数：2305843009213693951 (梅森素数)
    // 19位素数：4384957924686954497
    static boolean isPrime(long n) { //determines if n is a prime number
        int p[] = {2, 3, 5, 233, 331};
        int pn = p.length;
        long s = 0, t = n - 1;//n - 1 = 2^s * t
        while ((t & 1) == 0) {
            t >>= 1;
            ++s;
        }
        for (int i = 0; i < pn; ++i) {
            if (n == p[i]) {
                return true;
            }
            long pt = pow(p[i], t, n);
            for (int j = 0; j < s; ++j) {
                long cur = llMod(pt, pt, n);
                if (cur == 1 && pt != 1 && pt != n - 1) {
                    return false;
                }
                pt = cur;
            }
            if (pt != 1) {
                return false;
            }
        }
        return true;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
//        long r = 0;
//        a %= mod;
//        b %= mod;
//        while (b > 0) {
//            if ((b & 1) == 1) {
//                r = (r + a) % mod;
//            }
//            b >>= 1;
//            a = (a << 1) % mod;
//        }
//        return r;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        if (!oj) {
            System.setIn(new FileInputStream("in.txt"));
//            System.setOut(new PrintStream("out.txt"));
        }
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        long t = System.currentTimeMillis();
        solve();
        if (!oj) {
            pw.println("[" + (System.currentTimeMillis() - t) + "ms]");
        }
        pw.flush();
    }

    private static long[] anLong(int n) throws IOException {
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    private static String next(int len) throws IOException {
        int b = br.read();
        while (b == '\n' || b == ' ' || b == '\r' || b == '\t') {
            b = br.read();
        }
        char ch[] = new char[len];
        ch[0] = (char) b;
        int idx = 1;
        while (idx < len && (b = br.read()) != ' ' && b != '\n' && b != '\r' && b != '\t') {
            ch[idx++] = (char) b;
        }
        return String.valueOf(ch).trim();
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static long nextLong() throws IOException {
        return Long.parseLong(nextLine());
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

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