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 _577B_Modulo_Sum {
	public static void main(String[] args) {
	 Scanner leer = new Scanner(System.in);
	 int n = leer.nextInt();
	 int m = leer.nextInt();
	 boolean M[]= new boolean[m];
	 LinkedList<Integer> ms= new LinkedList<Integer>(),c;
	 ms.add(0);
	 for (int i = 0; i < n&&!M[0]; i++){
		 long a = leer.nextLong()%m;
		 c=(LinkedList<Integer>)ms.clone();
	     while(!c.isEmpty()&&!M[0]){
		     int  h= (c.remove()+(int)a)%m;
		     if(!M[h]){
		    	 M[h]=true;
		    	 ms.add(h);
		     }
		 }
	     if(!M[(int)a]){ms.add((int)a);M[(int)a]=true;}
	 }
     if(M[0])System.out.println("YES");
     else System.out.println("NO");
	}
}