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 B577 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		if(n >= m){
			System.out.println("YES");
			return;
		}
		
		int [] ayy = new int[n];
		boolean [][] lmao = new boolean[n][m];
		
		for(int i = 0; i<n; i++){
			ayy[i] = sc.nextInt();
			
		}
		
		

		
		for(int x = 0; x < n; x++){
			int num = ayy[x];
			int remainder = num % m;
			if(remainder == 0){
				System.out.println("YES");
				return;
			}
			lmao[x][remainder] = true;
			

			for(int y = 0; y < m; y++){
				if(x != 0 && lmao[x-1][y] == true){
					
					int lel = (num + y) % m;
					if(lel == 0){
						System.out.println("YES");
						return;
					}
					lmao[x][lel] = true;
					lmao[x][y] = true;
				}
				
				//System.out.print(lmao[x][y]);
			}
		}
		
		System.out.println("NO");
	}
}
