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.*;

public class Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int n = sc.nextInt();
              int m = sc.nextInt();
              int arr[] = new int[n];
              for(int i = 0 ; i < n ; i++)
              {
                     arr[i] = sc.nextInt();
                     
              }
              if(n > m)
              System.out.println("YES");
              
              else
              {
                     boolean is[][] = new boolean[n+1][m];
                     for(int i = 1 ; i <= n ; i++)
                     {
                            for(int j = 0 ; j < m ; j++)
                            {
                                   is[i][j] = is[i-1][j];
                                   
                            }
                            
                            int x = arr[i-1]%m;
                            for(int j = 0 ; j < m ; j++)
                            {
                                   if(j==0)
                                   is[i][x] = true;
                                   
                                   else
                                   {
                                          if(is[i-1][j])
                                          is[i][(j+x)%m] = true;
                                   }
                            }
                     }
                     
                    
                     if(is[n][0])
                     System.out.println("YES");
                     
                     else
                     System.out.println("NO");
              }
       }
}