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 n, m, ar[1010], dp[1010][1010][2];
bool fun(long long idx, long long cur_sum, bool taken) {
  if (cur_sum % m == 0 && taken) return 1;
  if (idx >= n) return 0;
  if (dp[idx][cur_sum][taken] != -1) return dp[idx][cur_sum][taken];
  bool ans1 = fun(idx + 1, (cur_sum % m), taken);
  bool ans2 = fun(idx + 1, (cur_sum + ar[idx]) % m, 1);
  return dp[idx][cur_sum][taken] = (ans1 || ans2);
}
void solve() {
  cin >> n >> m;
  if (n > m) {
    cout << "YES\n";
    return;
  }
  for (long long i = 0; i < n; i++) cin >> ar[i];
  memset(dp, -1, sizeof dp);
  bool yes = fun(0, 0, 0);
  if (yes)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
