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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class Codeforces577B {
	public static void main(String[] args) {
		try
		{
			BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
			StringTokenizer st = new StringTokenizer(f.readLine());
			int n = Integer.parseInt(st.nextToken());
			int m = Integer.parseInt(st.nextToken());
			st = new StringTokenizer(f.readLine());
			int[] array = new int[n];
			for(int i = 0; i < n; i++)
				array[i] = Integer.parseInt(st.nextToken())%m;
			boolean[][] subsequenceRemainder = new boolean[m][2]; 
			boolean found = false;
			subsequenceRemainder[array[0]][0] = true;
			if(subsequenceRemainder[0][0])
				found = true;
			for(int i = 1; i < n; i++)
			{
				for(int j = 0; j < m; j++)
				{
					if(subsequenceRemainder[(j - array[i] + m)%m][(i + 1)%2])
						subsequenceRemainder[j][i%2] = true;
					if(subsequenceRemainder[j][(i + 1)%2])
						subsequenceRemainder[j][i%2] = true;
				}
				subsequenceRemainder[array[i]][i%2] = true;
				if(subsequenceRemainder[0][i%2])
				{
					found = true;
					break;
				}
			}
			if(found)
				System.out.println("YES");
			else
				System.out.println("NO");
		}
		catch(IOException e)
		{
			System.out.println(e);
		}

	}

}
