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.ArrayList;
import java.util.Scanner;


public class CF_577B {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		if(n > m) {
			System.out.println("YES");
			return;
		}
		boolean[] canReach = new boolean[m];
		for(int i=0;i<n;i++) {
			int number = in.nextInt() % m;
			ArrayList<Integer> tmp = new ArrayList<>();
			for(int k=0;k<m;k++) {
				if(canReach[k]) {
					int val = (k+number)%m;
					if(!canReach[val]) {
						tmp.add(val);
					}
				}
			}
			for(int num: tmp)
				canReach[num] = true;
			canReach[number] = true;
		}
		if(canReach[0]) {
			System.out.println("YES");
		} else {
			System.out.println("NO");
		}
	}
}
