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 sudah[10000];
int n, m, x, y;
int sum;
int main() {
  memset(sudah, false, sizeof sudah);
  cin >> n >> m;
  if (n >= m) {
  }
  sum = 0;
  sudah[0] = true;
  vector<int> vi;
  while (n--) {
    cin >> x;
    for (int i = 0; i <= m; i++) {
      y = (i + x) % m;
      if (sudah[i]) {
        if (y == 0) {
          cout << "YES\n";
          return 0;
        }
        vi.push_back(y);
      }
    }
    for (int i = 0; i < vi.size(); i++) sudah[vi[i]] = true;
    vi.clear();
  }
  cout << "NO\n";
}
