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;

/**
 * Created by pallavi.v on 9/10/2015.
 */
public class Ideone {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] s = reader.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int m = Integer.parseInt(s[1]);
        boolean[] ch = new boolean[m];
        s = reader.readLine().split(" ");
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = (int) (Long.parseLong(s[i])%m);
        }
        for (int i = 0; i < n; i++) {
            int vl = a[i];
            boolean[] ch1 = new boolean[m];
            for(int j = 0; j < m; j++) {
                ch1[j] = ch[j];
            }
            for (int j = 0; j < m; j++) {
                if (ch[j]) ch1[(j+vl)%m] = true;
                if (ch1[0]) {
                    ch[0] = true;
                    break;
                }
            }
            ch = ch1;
            ch[vl] = true;
        }
        if (ch[0]) System.out.println("YES");
        else System.out.println("NO");
    }
}