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 n, k, a, b, i, j, f[1005];
vector<int> v;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  v.push_back(0);
  while (i < n) {
    cin >> a;
    j = 0;
    int sz = v.size();
    while (j < sz) {
      if ((v[j] + a) % k == 0) return cout << "YES", 0;
      if (f[(v[j] + a) % k] == 0) v.push_back((v[j] + a) % k);
      f[(v[j] + a) % k] = 1;
      j++;
    }
    i++;
  }
  i = 0;
  cout << "NO";
}
