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.Arrays;
import java.util.Scanner;

public class ModuloSum 
{
	static int m;
	
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		m = sc.nextInt();
		
		int[] reachable = new int[m+1];
		
		for(int i = 1; i <= n; i++)
		{
			int cur = sc.nextInt()%m;
			
			for(int j = 0; j < m+1; j++)
			{
				if(reachable[j] > 0 && reachable[(j+cur)%m] == 0 && reachable[j] != i)
				{
//					System.out.println(i + ") " + j + " -> " + ((j+cur)%m));
					reachable[(j+cur)%m] = i;
				}
			}
			
			reachable[cur] = i;
			if(reachable[0] > 0)
			{
				System.out.println("YES");
				return;
			}
//			System.out.println(Arrays.toString(reachable));
		}

		System.out.println("NO");
	}
}
