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.*;

public class ModuloSum {
    public static String solve(int[] nums, int m) {
        int n = nums.length;
        // pigeonhole principle
        // if n > m, then there must be a position where prefix_sum[0...i] = prefix_sum[0...j]
        // then the range in [i...j] is module m is zero
        if (n > m) return "YES";
        boolean[][] dp = new boolean[n][m];
        // dp[i][j] hold if at position i we could get module j
        for (int i = 0; i < n; ++i) {
            if (i == 0) {
                dp[i][nums[i] % m] = true;
            } else {
                for (int j = 0; j < m; ++j) {
                    int index = (j - nums[i] % m) >= 0 ? j - nums[i] % m : j + m - nums[i] % m;
                    dp[i][j] = dp[i-1][j] || dp[i-1][index];
                }
                dp[i][nums[i] % m] = true;
            }
        }

        if (dp[n-1][0] == true)
            return "YES";
        else
            return "NO";
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();
        int[] nums = new int[n];
        for (int i = 0; i < n; ++i)
            nums[i] = scan.nextInt();
        System.out.println(solve(nums, k));
    }
}