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.HashSet;
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;
		}
		HashSet<Integer> hs = new HashSet<Integer>(m);
		for (int i = 0; i < n; i++) {
			int number = in.nextInt() % m;
			if (number == 0) {
				System.out.println("YES");
				return;
			}
			HashSet<Integer> tmp = new HashSet<>();
			for (int k : hs) {
				int val = (k + number) % m;
				if (!hs.contains(val) && !tmp.contains(val)) {
					if (val == 0) {
						System.out.println("YES");
						return;
					}
					tmp.add(val);
				}
			}

			for (int num : tmp)
				hs.add(num);
			hs.add(number);
		}
		System.out.println("NO");

	}
}
