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.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class B577 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = in.nextInt(), m = in.nextInt();
		int[] count = new int[m];
	
		int[] a = new int[n + 1];
		
		for (int i=1; i<=n; i++) {
			a[i] = in.nextInt();
			a[i] %= m;
			count[a[i]] ++;
		}
		
		boolean ans = false;
		if (count[0] > 0 || n > m) {
			ans = true;
		} 
		
		for (int i=1; i<m; i++) {
			if (count[i] > 0 && count[m - i] > 0) {
				ans = true;
			}
			
			if (count[i] >= m) {
				ans = true;
			}
		}
		
		if (!ans) {
			boolean[][] dp = new boolean[n + 1][m];
			for (int i=1; i<=n; i++) {
				for (int j=0; j<m; j++) {
					if (dp[i-1][j]) {
						dp[i][j] = true;
						dp[i][(j + a[i]) % m] = true;
					}
				}
				dp[i][a[i]] = true;
			}
			if (dp[n][0]) {
				ans = true;
			}
		}
		
		if (ans) {
			out.println("YES");
		} else {
			out.println("NO");
		}
		
		out.close();
	}

}
