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;
long long a[(long long)1e6 + 10], ca[1010][1010];
int main() {
  long long n, m;
  cin >> n >> m;
  bool res = false;
  if (n > m)
    res = true;
  else {
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      a[i] %= m;
    }
    ca[0][a[0] % m] = 1;
    for (long long i = 1; i < n; i++) {
      ca[i][a[i] % m] = 1;
      for (long long j = 0; j < m; j++) {
        ca[i][j] =
            ca[i][j] | ca[i - 1][(j - (a[i] % m) + m) % m] | ca[i - 1][j];
      }
    }
    res = ca[n - 1][0];
  }
  cout << (res ? "YES" : "NO");
  return 0;
}
