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 B {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] nums = new int[n];
		for (int i = 0; i < n; i++)
			nums[i] = sc.nextInt();
		sc.close();

		if (n > m)
			System.out.println("YES");
		else {
			boolean[][] dp = new boolean[n][m];
			boolean found = dp[0][0];
			for (int i = 0; i < n && !found; i++) {
				dp[i][nums[i] % m] = true;
				if (i < n - 1)
					for (int j = 0; j < m; j++)
						if (dp[i][j]) {
							dp[i + 1][j] = true;
							dp[i + 1][(j + nums[i + 1]) % m] = true;
						}
				if (dp[i][0])
					found = true;
			}
			System.out.println((found ? "YES" : "NO"));
		}
	}
}
