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 cf_577B {
	static boolean[][] memo;
	static int[] arr;
	static boolean works;
	static int n,m;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		m = in.nextInt();
		arr = new int[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = (int) (in.nextLong() % m);
			if(arr[i]==0){
				System.out.println("YES");
				return;
			}
		}
		if(n>m){
			System.out.println("YES");
		} else {
			memo = new boolean[n][m];
			works = false;
			go(0,-1);
			if(works)System.out.println("YES");
			else System.out.println("NO");
		}
	}
	private static void go(int i, int sum) {
		//System.out.println(i + " " + sum);
		if(sum==0){
			works = true;
			return;
		}
		if(i>=memo.length || works || (sum!=-1 && memo[i][sum]))return;
		if(sum!=-1)memo[i][sum] = true;
		go(i+1, sum);
		int t = (sum==-1?arr[i]:(sum+arr[i])) % m;
		go(i+1, t);
	}
}
