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.util.Scanner;

public class CodeForces319B {
    public static void main(String[] args) {
        Scanner scn=new Scanner(System.in);
        int N=scn.nextInt();
        int M=scn.nextInt();
        long[] A=new long[N];
        for(int i=0;i<N;i++){
            A[i]=scn.nextLong();
        }
        if(N>M){
            System.out.println("YES");
        }
        else{
            dp=new int[N][M];
            if(cal(A, 0, 0, M)==1){
                System.out.println("YES");
            }
            else{
                System.out.println("NO");
            }
        }


    }

    private static int cal(long[] A, int i, int sum,  int m) {
        if(i==A.length){
            return 2;
        }
        if((sum+A[i])%m==0){
            dp[i][sum]=1;
            return 1;
        }
        if(dp[i][sum]!=0){
            return dp[i][sum];
        }

        if(cal(A, i+1, sum, m)==1||cal(A, i+1, (int) ((sum+A[i])%m), m)==1){
            dp[i][sum]=1;
        }
        else{
            dp[i][sum]=2;
        }
        return dp[i][sum];
    }

    private static int[][] dp;
}
