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.*;
import java.io.*;
import java.math.*;
public class Test{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		int m = in.nextInt();
		int[] arr = new int[n];
		boolean found = false;
		for(int i=0;i<n;++i){
			arr[i] = in.nextInt()%m;
			found = found | (arr[i]==0);	
		}	
		if(n>=m || found){
			out.println("YES");
			out.close();
			return;
		}
		boolean dp[][] = new boolean[n+1][m+1];
		for(int i=1;i<=n;++i){
			dp[i][arr[i-1]] = true;
			for(int j=0;j<m;++j){
				dp[i][j] |= dp[i-1][j];
				dp[i][j] |= dp[i-1][(j-arr[i-1]+m)%m];
			}
		}
		out.println(dp[n][0] ? "YES" : "NO");		
		out.close();
	}
}
