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.*;
public class Solution {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	//Main
	public static void main(String args[]) {
		int test=1;
		//test=sc.nextInt();
		while(test-->0) {
			int n=sc.nextInt(),m=sc.nextInt(),a[]=new int[n];
			for(int i=0;i<n;i++) a[i]=sc.nextInt()%m;
			if(n>m) {
				out.println("YES");
				continue;
			}
			int dp[][]=new int[n][m];
			for(int i=0;i<n;i++) {
				for(int j=0;j<m;j++) {
					if(i>0) dp[i][j]=dp[i-1][j] | dp[i-1][(j-a[i]+m)%m];
				}
				dp[i][a[i]]=1;
			}
			out.println(dp[n-1][0]==1?"YES":"NO");
			
		}
		out.flush();
		out.close();
	}
}
