You are given a sequence of numbers a1, a2, ..., an, and a number m.

Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

Input

3 5
1 2 3


Output

YES


Input

1 6
5


Output

NO


Input

4 6
3 1 1 3


Output

YES


Input

6 6
5 5 5 5 5 5


Output

YES

Note

In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.
//package codechef;

import java.util.Scanner;

/**
 * Created by nitin.s on 13/09/15.
 */
public class ModuloSum {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        boolean res = false;
        for(int i = 0; i < n; ++i) {
            a[i] = sc.nextInt();
            a[i] %= m;
            if(a[i] % m == 0) {
                res = true;
            }
        }
        if(n >= m) {
            System.out.println("YES");
            return;
        }
        if(res) {
            System.out.println("YES");
            return;
        }
        boolean[][] can = new boolean[n + 1][m + 1];
        for(int i = 1; i <= n && !res; ++i) {
            can[i][a[i - 1]] = true;
            for(int j = 0; j < m; ++j) {
                if(can[i - 1][j]) {
                    can[i][j] = true;
                    int temp = (a[i - 1] + j) % m;
                    can[i][temp] = true;
                    if(j == 0 || temp == 0) {
                        res = true;
                        break;
                    }
                }
            }
        }

        if(res) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}
