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 mod = 1e9 + 7;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
bool poss[2][1000];
void solve() {
  int n, m;
  cin >> n >> m;
  if (n > m) {
    cout << "YES\n";
    return;
  }
  int i, t, j;
  for (i = 0; i < n; i++) {
    cin >> t;
    poss[1][t % m] = 1;
    for (j = 0; j < m; j++)
      if (poss[0][j] == 1) poss[1][(j + t) % m] = 1;
    for (j = 0; j < m; j++) poss[0][j] = max(poss[1][j], poss[0][j]);
  }
  if (poss[0][0] == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
