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

public class Main {
    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
        
        int n = in.nextInt();
        int m = in.nextInt();
        
        boolean[][] dp = new boolean[1050][2];
        
        for (int i = 1; i <= n; i++) {
            int curNum = in.nextInt();
            curNum = curNum % m;
            int prev = 0;
            int cur = 1;
            if (i % 2 == 0) {
                prev = 1;
                cur = 0;
            }
            
            dp[curNum][cur] = true;
            for (int j = 0; j < m; j++) {
                if (dp[j][prev] == true) {
                    int sum = (j + curNum) % m;
                    dp[sum][cur] = true;
                    dp[j][cur] = true;
                }
            }
            
            if (dp[0][cur] == true) {
                System.out.println("YES");
                return;
            }
        }
        
        System.out.println("NO");   
    }
}