You are given an array a_1, a_2, ... , a_n and two integers m and k.

You can choose some subarray a_l, a_{l+1}, ..., a_{r-1}, a_r. 

The cost of subarray a_l, a_{l+1}, ..., a_{r-1}, a_r is equal to ∑_{i=l}^{r} a_i - k ⌈ (r - l + 1)/(m) ⌉, where ⌈ x ⌉ is the least integer greater than or equal to x. 

The cost of empty subarray is equal to zero.

For example, if m = 3, k = 10 and a = [2, -4, 15, -3, 4, 8, 3], then the cost of some subarrays are:

  * a_3 ... a_3: 15 - k ⌈ 1/3 ⌉ = 15 - 10 = 5; 
  * a_3 ... a_4: (15 - 3) - k ⌈ 2/3 ⌉ = 12 - 10 = 2; 
  * a_3 ... a_5: (15 - 3 + 4) - k ⌈ 3/3 ⌉ = 16 - 10 = 6; 
  * a_3 ... a_6: (15 - 3 + 4 + 8) - k ⌈ 4/3 ⌉ = 24 - 20 = 4; 
  * a_3 ... a_7: (15 - 3 + 4 + 8 + 3) - k ⌈ 5/3 ⌉ = 27 - 20 = 7. 



Your task is to find the maximum cost of some subarray (possibly empty) of array a.

Input

The first line contains three integers n, m, and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 10, 1 ≤ k ≤ 10^9).

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9).

Output

Print the maximum cost of some subarray of array a.

Examples

Input


7 3 10
2 -4 15 -3 4 8 3


Output


7


Input


5 2 1000
-13 -4 -9 -20 -11


Output


0
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.security.cert.CollectionCertStoreParameters;
import java.util.*;
import java.io.BufferedReader;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(1, in, out);
        out.close();
    }

    static class Solver {
        static int n, m, k;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            n = in.nextInt();
            m = in.nextInt();
            k = in.nextInt();
            long[] p = new long[n + 1];
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
                p[i + 1] = p[i] + a[i];
            }

            long result = Long.MIN_VALUE;
            long[] mins = new long[m];
            long[] psums = new long[m];

            for (int r = 0; r < n; r++) {
                long lastSum = 0;
                for (int last = 0; last < Math.min(m, r + 1); last++) {
                    int end = r - last;
                    long endSum = p[end + 1] - p[Math.max(0, end + 1 - m)] - k;

                    if (last == 0) {
                        long newPartialSum = psums[end % m] + endSum;
                        psums[end % m] = newPartialSum;
                        mins[end % m] = Math.min(mins[end % m], newPartialSum);
                    }

                    long best = psums[end % m] - mins[end % m];
                    result = Math.max(result, best + lastSum - k * (last > 0 ? 1 : 0));

                    if (r - last >= 0) {
                        lastSum += a[r - last];
                    }
                }
            }

            out.println(result);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

    }
}
 