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.*;
public class Abc
{
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n,m;
		n=in.nextInt();
		m=in.nextInt();
		int ar[]=new int[n];
		for(int i=0;i<n;i++)
		{
			ar[i]=in.nextInt();
		}
		if(n>=m)
		{
			System.out.println("YES");
			return;
		}
		int can[][]=new int[n+1][m];
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(can[i][j]==1)
				{
					can[i+1][(ar[i]+j)%m]=1;	
					can[i+1][j]=1;
				}
				if(ar[i]%m==j)
				{
					can[i][j]=1;
					can[i+1][j]=1;
				}
			}
		}
		//print(can);
		if(can[n][0]==1)
		{
			System.out.println("YES");
		}
		else
		{
			System.out.println("NO");
		}
	}
	public static void print(int ar[][])
	{
		for(int i=0;i<ar.length;i++)
		{
			for(int j=0;j<ar[0].length;j++)
			{
				System.out.print(ar[i][j]+" ");
			}
			System.out.println();
		}
	}
}