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;
bool can[1005][1005];
long long a[1000002];
int main() {
  int n, m;
  cin >> n >> m;
  if (n > m) {
    cout << "YES" << endl;
    return 0;
  } else {
    long long mi = INT_MAX;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] < mi) mi = a[i];
    }
    memset(can, false, sizeof(can));
    for (int i = 1; i <= n; i++) {
      can[i][a[i] % m] = true;
      for (int j = 0; j <= m; j++) {
        if (can[i - 1][j]) {
          can[i][j] = true;
          can[i][((j % (m)) + ((a[i]) % (m))) % (m)] = true;
        }
      }
    }
    bool ans = false;
    for (int i = 1; i <= n; i++) {
      if (can[i][0]) {
        ans = true;
        break;
      }
    }
    if ((ans) || (mi == 0))
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}
