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.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;


public class Solution {
	
	Scanner cin = new Scanner(System.in);
	PrintWriter cout = new PrintWriter(System.out);
	
	int n, m;
	byte prev[], cur[];
	int x;
	
	private void run() {
		n = cin.nextInt();
		m = cin.nextInt();
		
		if (n > m) {
			cout.println("YES");
		} else {
			prev = new byte[m];
			cur = new byte[m];
			
			for (int i = 0; i < m; ++i) {
				prev[i] = 0;
				cur[i] = 0;
			}
			
			for (int i = 0; i < n; ++i) {
				x = cin.nextInt();
				
				x = x % m;
				cur[x] = 1;
				for (int j = 0; j < m; ++j) {
					if (prev[j] == 0) {
						continue;
					}
					cur[j] = 1;
					cur[(j + x) % m] = 1;
				}
				
				for (int j = 0; j < m; ++j) {
					prev[j] = cur[j];
					//cout.print(prev[j] + " ");
				}
				//cout.println();
			}
			if (prev[0] == 1) {
				cout.println("YES");
			} else {
				cout.println("NO");
			}
		}
		
		cin.close();
		cout.close();
	}
	
	public static void main(String[] args) {
		Solution solution = new Solution();
		solution.run();
	}
}
