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.
//created by Whiplash99
import java.io.*;
import java.util.*;
public class A
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N;

        String[] s=br.readLine().trim().split(" ");
        N=Integer.parseInt(s[0]);
        int M=Integer.parseInt(s[1]);

        int[] a=new int[N];
        s=br.readLine().trim().split(" ");
        for(i=0;i<N;i++) a[i]=(Integer.parseInt(s[i]))%M;

        if(N>M) //Pigeonhole Principle
        {
            System.out.println("YES");
            System.exit(0);
        }

        boolean[][] dp=new boolean[N+1][M+1];
        for(i=0;i<=N;i++) dp[i][0]=true;

        for(i=1;i<=N;i++) //Subset-Sum DP
        {
            for(int j=1;j<=M;j++)
            {
                if(a[i-1]==0) dp[i][j]=true;

                dp[i][j]|=dp[i-1][j];
                if(a[i-1]>j) dp[i][j]|=dp[i-1][j-a[i-1]+M];
                else dp[i][j]|=dp[i-1][j-a[i-1]];
            }
        }
        System.out.println(dp[N][M]?"YES":"NO");
    }
}