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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  if (n > m) {
    cout << "YES" << endl;
    return 0;
  }
  vector<int> numbers(n + 1);
  numbers[0] = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> numbers[i];
    numbers[i] %= m;
    if (numbers[i] == 0) {
      cout << "YES" << endl;
      return 0;
    }
  }
  vector<vector<int>> dp(n + 1);
  for (int i = 0; i <= n; ++i) {
    dp[i].resize(m + 1);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      int res = (j - numbers[i]) % m;
      if (res < 0) res += m;
      if (dp[i - 1][j] % m == j % m && dp[i - 1][j] != 0)
        dp[i][j] = dp[i - 1][j];
      else {
        if ((dp[i - 1][res] + numbers[i]) % m == j % m)
          dp[i][j] = dp[i - 1][res] + numbers[i];
        else
          continue;
      }
    }
  }
  if (dp[n][m] % m == 0 && dp[n][m] != 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
