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 MAXN = 1e6 + 100;
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    int temp, ans = 0;
    set<int> st, _st;
    st.insert(0);
    set<int>::iterator iter;
    for (int i = 0; i < n; i++) {
      scanf("%d", &temp);
      if (ans) continue;
      for (iter = st.begin(); iter != st.end(); iter++) {
        if ((*iter + temp) % m == 0) {
          ans = 1;
          break;
        }
        _st.insert((*iter + temp) % m);
      }
      st.insert(_st.begin(), _st.end());
      _st.clear();
    }
    if (ans)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
