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.IOException;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
import java.util.stream.Collectors;
import java.util.stream.Stream;

/**
 * Created by Tejas on 16-07-2018.
 */
public class Main {

    public static void main(String[]args)throws IOException{
        Scanner scanner=new Scanner(System.in);
        HashSet<Integer> set=new HashSet<>();
        int N=scanner.nextInt();
        int M=scanner.nextInt();

        for (int i = 0; i < N && !set.contains(0); i++) {
            int x = scanner.nextInt() % M;
            Set<Integer> newSet= Stream.concat(Stream.of(x % M),
                        set.stream().map(y -> (y + x) % M))
                        .collect(Collectors.toSet());
                set.addAll(newSet);
        }

        if(set.contains(0))
            System.out.println("YES");
        else System.out.println("NO");

    }




}