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(NULL);
  long int n, m;
  cin >> n >> m;
  long int a[1000005];
  vector<int> v;
  int vis[1005];
  int flg = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = a[i] % m;
    if (a[i] == 0) flg = 1;
  }
  if (flg == 1) {
    cout << "YES\n";
    return 0;
  }
  for (int i = 0; i < n && flg == 0; i++) {
    if (v.empty()) {
      vis[a[i]] = 1;
      v.push_back(a[i]);
    } else {
      int len = v.size();
      for (int j = 0; j < len; j++) {
        int nw = (v[j] + a[i]) % m;
        if (vis[nw] == 0) {
          vis[nw] = 1;
          v.push_back(nw);
        }
      }
      if (vis[a[i]] == 0) {
        vis[a[i]] = 1;
        v.push_back(a[i]);
      }
      if (vis[0] == 1) flg = 1;
    }
  }
  if (flg)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
