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() {
  if (0 == 1) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int n, m;
  int t, i;
  int luu[1100];
  bool check = false;
  cin >> n >> m;
  luu[0] = 1;
  int luutmp[1100];
  for (int i = 1; i <= n; i++) {
    int value;
    cin >> value;
    value %= m;
    for (int j = 0; j <= m - 1; j++) {
      luutmp[j] = 0;
    }
    for (int j = 0; j <= m - 1; j++)
      if (luu[j] > 0) {
        int newValue = (j + value) % m;
        luutmp[newValue] = 1;
      }
    if (luutmp[0] == 1) {
      check = true;
      break;
    }
    for (int j = 0; j <= m - 1; j++) {
      if (luutmp[j] == 1) luu[j] = 1;
    }
  }
  if (check == true) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  return 0;
}
