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 JavaApplication54 {

   
    public static void main(String[] args) {
   int x,n,m;
     Scanner in; 
        in = new Scanner(System.in);
       n=in.nextInt();
       m=in.nextInt();
       int []a=new int[10000];
       int []b=new int[10000];
       for(int i = 0 ; i<10000;i++)
           a[i]=b[i]=0;
       
       for(int i =1; i <=n;i++)
       {
       x=in.nextInt();
       for(int j=0;j<m;j++)
       {
           a[j]=b[j];
      }
       for(int j = 0; j<m ; j++)
       {
       if(a[j]==1)
       {
       b[(j+x)%m]=1;
       }
       }
       b[x%m]=1;
       if(b[0]==1)
       {System.out.println("YES"); return;}
       
       
       }
        System.out.println("NO");
       }
       
     
  
   
    }