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;

public class Main {

  public static boolean[][] dp = new boolean[2][1000];

  public static void main(String[] args) {
    // TODO Auto-generated method stub

    Scanner scn = new Scanner(System.in);

    int N = scn.nextInt();
    int M = scn.nextInt();

    int[] A = new int[N];

    for (int j = 0; j < M; j++) {
      dp[0][j] = false;
      dp[1][j] = false;
    }

    for (int i = 0; i < N; i++) {
      A[i] = scn.nextInt() % M;
    }

    for (int i = 1; i <= N; i++) {

      for (int j = 0; j < M; j++) {

        if (dp[i % 2][j]) {
          dp[(i + 1) % 2][(j + A[i - 1]) % M] = true;
          dp[(i + 1) % 2][j] = true;
        }
      }

      dp[(i + 1) % 2][A[i - 1]] = true;

      if (dp[(i + 1) % 2][0]) {
        dp[N % 2][0] = true;
        break;
      }
    }

    System.out.println(dp[N % 2][0] ? "YES" : "NO");
  }
}
