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;


/**
 * Created by bobyk on 22/08/15.
 */
public class Main {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        long[] a = new long[2000000];
        for (int i = 0; i < n; i++){
            long x = sc.nextLong();
            a[i] = x;
        }

        if (n > m){
            int[] s = new int[2000000];
            long[] k = new long[2000000];
            boolean ok = false;
            for (int i = 0; i < n; i++){
                if (i > 0) s[i] = (int) ((s[i-1] - a[i]) % m);
                else s[i] = (int)(a[i])%m;
                while (s[i] < 0)
                    s[i]+=m;
                k[s[i]]++;
                if (k[s[i]] > 1) ok = true;
            }
            if (ok) System.out.println("YES");
            else System.out.println("NO");
        }
        else {
            int[][] dp = new int[3000][3000];
            for (int i = 0; i < n; i++){
                for (int j = 0; j < m; j++){
                    dp[i][j] = -1;
                }
            }
            dp[0][0] = 0;
            for (int i = 0; i < n; i++){
                for (int residue = 0; residue < m; residue++){
                    if (dp[i][residue] != -1){
                      //  System.out.println(i + " " + residue);
                        dp[i+1][(int)((residue+a[i])%m)] = Math.max(dp[i+1][(int)((residue+a[i])%m)], dp[i][residue] + 1);
                        dp[i+1][residue] = Math.max(dp[i+1][residue], dp[i][residue]);
                    }
                }
            }
            if (dp[n][0] > 0) System.out.println("YES");
            else System.out.println("NO");
        }

    }

}
