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.Scanner;

public class ModuloSum {

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);

		int n = scanner.nextInt();
		int target = scanner.nextInt();

		boolean mods[] = new boolean[target];
		int nums[] = new int[target];
		int count = 0;

		for (int i = 0; i < n; ++i) {
			int num = scanner.nextInt() % target;
			if (num == 0) {
				System.out.println("YES");
				scanner.close();
				return;
			}

			for (int j = count; j >= 0; --j) {
				int newMod = (nums[j] + num) % target;
				if (newMod == 0) {
					System.out.println("YES");
					scanner.close();
					return;
				}

				if (!mods[newMod]) {
					nums[count++] = newMod;
					mods[newMod] = true;
				}
			}
		}

		System.out.println("NO");
		scanner.close();

	}

}
