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.Arrays;

public class ModuloSum {
	static int[] numbers;
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] nm = br.readLine().split(" ");
		int n = Integer.parseInt(nm[0]);
		int m = Integer.parseInt(nm[1]);
		numbers = new int[n];
		String[] numbersString = br.readLine().split(" ");
		for(int i = 0; i < n; i++)
			numbers[i] = Integer.parseInt(numbersString[i]);
		if(n > m){
			System.out.println("YES");
			return;
		}
		int[][] dp = new int[n][m+1];
		if(isThereSubsequence(n-1, m, dp, m)) System.out.println("YES");
		else System.out.println("NO");
	}
	// existe alguien que sume 0 modulo m ?
	private static boolean isThereSubsequence(int n, int m, int[][] dp,int mod) {
		if(n == 0) return (numbers[0]%mod == 0 || numbers[0]%mod == m);
		if(numbers[n]%mod == 0 || numbers[n]%mod == m) return true;
		if(dp[n-1][m] == 0) dp[n-1][m] = isThereSubsequence(n-1, m, dp,mod) ? 1 : -1;
		if(dp[n-1][m] == 1) return true;
		if(dp[n-1][(mod + m-numbers[n]%mod)%mod] == 0) dp[n-1][(mod + m-numbers[n]%mod)%mod] = isThereSubsequence(n-1, (mod + m-numbers[n]%mod)%mod, dp,mod) ? 1 : -1;
		if(dp[n-1][(mod + m-numbers[n]%mod)%mod] == 1) return true;
		return false;
	}
//	private static boolean isThereSubsequence(int n, int sum, int m, int[][] dp) {
//		if(n == 0) return numbers[0]%m == sum%m;
//		if(numbers[n]%m == sum%m) return true;
//		
//		if(dp[n-1][sum] == 0) dp[n-1][sum] = isThereSubsequence(n-1, sum, m, dp) ? 1 : -1;
//		if(dp[n-1][sum] == 1) return true;
//		
//		if(sum-numbers[n]%m < 0) return false;
//		if(dp[n-1][sum-numbers[n]%m] == 0) dp[n-1][sum-numbers[n]%m] = isThereSubsequence(n-1, sum-numbers[n]%m, m, dp) ? 1 : -1;
//		if(dp[n-1][sum-numbers[n]%m] == 1) return true;
//		
//		return false;
//	}

}
