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;
const int Maxm = 1e3 + 100;
int cnt[Maxm];
bool mk[Maxm], mk2[Maxm], flag[Maxm], res;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int tmp;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &tmp);
    tmp %= m;
    if (cnt[tmp] < m) cnt[tmp]++;
  }
  mk2[0] = mk[0] = true;
  for (int i = 0; !res && i < Maxm; ++i) {
    for (int j = 0; !res && j < Maxm; ++j)
      if (mk[j] && !flag[j]) {
        for (int k = 1; !res && k <= cnt[i]; ++k) {
          if ((k * i + j) % m == 0) res = true;
          if (k < cnt[i]) {
            mk2[(k * i + j) % m] = true;
          } else
            mk2[(k * i + j) % m] = true;
        }
      }
    for (int j = 0; j < Maxm; ++j) {
      mk[j] = mk2[j];
      flag[j] = false;
    }
  }
  if (res)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
