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 int mod = 1e9 + 7;
bool solve(vector<long long int> &v, long long int k) {
  if (v.size() > k) return 1;
  bool dp[k];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < v.size(); i++) {
    if (dp[0]) return 1;
    bool temp[k];
    memset(temp, 0, sizeof(temp));
    for (int j = 0; j < k; j++) {
      if (dp[j]) {
        temp[(j + v[i]) % k] = 1;
      }
    }
    for (int j = 0; j < k; j++)
      if (temp[j]) dp[j] = 1;
    dp[v[i] % k] = 1;
  }
  return dp[0];
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  long long int t, n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  if (solve(v, k))
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
  return 0;
}
