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.


////////////////////---------------------------SHUBHAM CHAUDHARI-------------------------------///////////////////////


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


public class Main {

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


    }

}
