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 Main {
	
	private static int[] input = new int[1000010];
	private static boolean[] dp1 = new boolean[2000];
	private static boolean[] dp2 = new boolean[2000];

	public static void main(String[] args) throws Exception {		
	
		Scanner scan = new Scanner(System.in);
		int i , j , n , m;
		n = scan.nextInt();
		m = scan.nextInt();
		for (i = 0;i < n;i ++) {
			input[i] = scan.nextInt();
		}
	
		if (n > m) {
			System.out.println("YES");
		} else {
			for (i = 0;i < n;i ++) {
				for (j = 0;j < m;j ++) {
					dp2[j] = dp1[j];
				}
				int index = input[i] % m;
				for (j = 0;j < m;j ++) {
					if (dp1[j] || j == 0) {
						int next = (j + index) % m;
						dp2[next] = true;
					}
				}
				for (j = 0;j < m;j ++) {
					dp1[j] = dp2[j];
					dp2[j] = false;
				}
			}
			if (dp1[0]) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}
		
	}
	
	
}








