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>
using namespace std;
long long int f = 0, x, j, i, n, m, dp[1005] = {0}, t, temp[1005] = {0};
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> x;
    x = x % m;
    if (x == 0) {
      f = 1;
    } else if (f == 0) {
      for (j = 1; j < m; j++) {
        if (dp[j] != 0) {
          t = (j + x) % m;
          if (t == 0)
            f = 1;
          else
            temp[t] = 1;
        }
      }
      for (j = 1; j < m; j++) {
        if (temp[j] == 1) dp[j] = temp[j];
        temp[j] = 0;
      }
      dp[x] = 1;
    }
  }
  f ? cout << "YES" : cout << "NO";
  return 0;
}
