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.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;


public class A {

    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);
//        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
//        String[] ar = scan.readLine().split(" ");
        int n = in.nextInt();
        int m = in.nextInt();
        int a[] = new int[m];
        for (int i = 1;i <= n; i++) {
            int x = in.nextInt() % m;
            for (int j = 0; j < m; j++)
                if (a[j] > 0 && a[j] != i && a[(j + x) % m] == 0)
                    a[(j + x) % m] = i;            
            a[x] = i;
//            for (int j = 0; j < m; j++)
//                System.out.print(a[j] + " ");
//            System.out.println();
            if (a[0] != 0) {
                System.out.println("YES");
                return;
            }
        }
        System.out.println("NO");
        
    }
}