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 Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		int m=s.nextInt();
		
		long[] arr=new long[n];
		
		for(int i=0;i<n;i++)
		{
			arr[i]=s.nextLong();
		}
		
		boolean ans=find(arr,n,m);
		
		if(ans)
			System.out.println("YES");
		else
			System.out.println("NO");
		
	}
	
	public static boolean find(long[] arr,int n,int m)
	{
		if(n>m)
			return true;
		
		boolean[] dp=new boolean[m+1];
		
		for(int i=0;i<n;i++)
		{
			if(dp[0])
				return true;
			
			boolean[] temp=new boolean[m+1];
			
			for(int j=0;j<=m;j++)
			{
				if(dp[j])
				{
					if(dp[(int)(j+arr[i])%m]==false)
					{
						temp[(int)(j+arr[i])%m]=true;
					}
				}
			}
			
			for(int j=0;j<=m;j++)
			{
				if(temp[j])
				{
					dp[j]=true;
				}
			}
			
			dp[(int)(arr[i]%m)]=true;
			
		}
		
		
		
		return dp[0];
	}
	
}