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.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    public static void main(String[] args) throws IOException
    {
     
      Scanner vicky=new Scanner(System.in);
      int n=vicky.nextInt();
      int m=vicky.nextInt();
      int [] ar=new int[n];
      int [] br=new int [m];

      for(int i=0;i<n;i++)
      {
          ar[i]=vicky.nextInt();
          
          if(ar[i]%m==0){System.out.println("YES");return;}
          ar[i]%=m;
          if(br[m-ar[i]]==1)
          {System.out.println("YES");return;}
          int [] t=new int [m];
          for(int p=0;p<m;p++)
              t[p]=br[p];
          t[ar[i]]=1;
          for(int j=m-1;j>=1;j--)
          {
              if(br[j]!=0 )t[(j+ar[i])%m]=1;
          }
          for(int p=0;p<m;p++)
              br[p]=t[p]; 
      }
      System.out.println("NO");
      
    }
    
}