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.*;
import java.lang.*;

public class Rextester{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        StringTokenizer st1 = new StringTokenizer(br.readLine());
        br.close();
        if(n>m){
            System.out.println("Yes");
            return;
        }
        int[] array = new int[n];
        for(int i=0;i<n;i++){
            array[i]=(Integer.parseInt(st1.nextToken())%m);
        }
        boolean[][] dp = new boolean[n][m];
        boolean[][] done = new boolean[n][m];
        findAns(dp,done,n-1,m,0,array);
        for(int i=n-2;i>=0;i--){
            if(dp[i][(m-array[i+1])%m]){
                System.out.println("Yes");
                return;
            }
        }
        if(n==1){
            if(array[0]%m==0){
                System.out.println("Yes");
                return;
            }
        }
        System.out.println("No");
        
    }
    static boolean findAns(boolean[][] dp,boolean[][] done,int n,int m,int mod,int[] array){
        if(n<0 && mod==0){
            return true;
        }
        else if(n<0){
            return false;
        }
        if(!done[n][mod]){
            dp[n][mod]=findAns(dp,done,n-1,m,mod-array[n]<0?mod-array[n]+m:mod-array[n],array)|findAns(dp,done,n-1,m,mod,array);
            done[n][mod]=true;
        }
        return dp[n][mod];
    }
}
        
        
            