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.*;
import java.io.PrintWriter;

public class test
{
	public static void main(String[] arg)
	{
		new test();
	}
	public test()
	{
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		int m = in.nextInt();
		boolean[] used = new boolean[m];
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		for(int i = 0; i < n; i++)
		{
			if(m == 0) break;
			int temp = in.nextInt()%m;
			int size = q.size();
			if(!used[temp])
			{
				q.add(temp);
				used[temp] = true;
			}
			while(size-- > 0)
			{
				int num = q.poll();
				q.add(num);
				int ele = (num+temp)%m;
				if(used[ele]) continue;
				used[ele] = true;
				q.add(ele);
			}
			if(used[0]) break;
		}
		out.println(used[0]?"YES":"NO");
		in.close(); out.close();
	}
}