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 B 
{

	static int n , m ;
	static Boolean [][][] memo ; 
	static int [] a ; 
	
	static boolean dp(int idx , int sum, int taken)
	{
		if(sum == 0 && taken == 1)
			return true ; 
		
		if(idx == n)
			return false ; 
		
		if(memo[idx][sum][taken] != null)
			return memo[idx][sum][taken];
		
		return memo[idx][sum][taken] = dp(idx + 1, (sum + a[idx]) % m, 1 ) | dp(idx + 1, sum, taken);
	}
	
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt(); m = sc.nextInt();
		if(n > m)
		{
			System.out.println("YES");
			return ;
		}
		
		a = new int [n];
		memo = new Boolean[n][m][2];
		for(int i = 0 ; i < n ;i++)
			a[i] = sc.nextInt();
		
		System.out.println(dp(0, 0, 0) ? "YES":"NO");
		

	}


}
