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 Practice1.CF577;

import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class CF577B {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int m = s.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        if(n > m){
            System.out.println("YES");
        }else{
            boolean[][] mat = new boolean[n+1][m];
            for (int i = 0; i < n; i++) {
                int val = arr[i];
                mat[i+1][val % m] = true;
                for (int j = 0; j < m; j++) {
                    if(mat[i][j]){
                        mat[i+1][j] = true;
                        mat[i+1][(j+val) % m] = true;
                    }
                }
            }
            if(mat[n][0]){
                System.out.println("YES");
            }else{
                System.out.println("NO");
            }

        }
    }


}
