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 modulosum {
	public static Scanner scn = new Scanner(System.in);
	

	public static void main(String[] args) {

		int n = scn.nextInt();
		int m = scn.nextInt();
		if (n >= m) {
			for (int i = 0; i < n; i++) {
				scn.nextInt();
			}
			System.out.println("YES");
		} else {
			int[] a=new int[n+1];
			for(int i=1;i<=n;i++){
				a[i]=scn.nextInt();
			}
			solve(a,n,m);
		}

	}

	

	public static void solve(int[] a, int n, int m) {

		int[][] dp = new int[m+1][m+1];
		dp[0][0] = 1;
		for (int i = 1; i <= n; i++) {
			int t = a[i];
			for (int j = 0; j < m; j++) {
				if (dp[j][i - 1] != 0) {
					dp[j][i]++;
					dp[(j + t) % m][i]++;
				}
			}

		}

		for (int i = 1; i < m; i++) {
			if (dp[0][i] > 1) {
				System.out.println("YES");
				return;
			}
		}
		System.out.println("NO");
		return;
		
	}

}
