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.
#include <bits/stdc++.h>
const int MAX_M_SIZE = 1005;
int mod_tag[MAX_M_SIZE];
int res_tag[MAX_M_SIZE];
int main() {
  int n, m;
  int num_tmp;
  scanf(" %d %d ", &n, &m);
  {
    memset(mod_tag, 0, sizeof(mod_tag));
    memset(res_tag, 0, sizeof(res_tag));
    int yes_tag = 0;
    int round = 1;
    for (int i = 1; i <= n; i++) {
      scanf(" %d ", &num_tmp);
      num_tmp %= m;
      for (int j = 1; j < m; j++) {
        if (res_tag[j] < round && !res_tag[(j + num_tmp) % m] && res_tag[j]) {
          res_tag[(j + num_tmp) % m] = round;
        }
      }
      res_tag[num_tmp] = round;
      round++;
      if (res_tag[0]) {
        yes_tag = 1;
        break;
      }
    }
    if (yes_tag) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
