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;
void print(vector<long long> v) {
  long long n = v.size();
  for (long long i = 0; i < n; i++) {
    cout << v[i] << " ";
  }
  cout << '\n';
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b != 0) {
    if (b & 1) {
      res *= a;
    }
    a = (a * a);
    b = (b / 2);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  if (n > m) {
    cout << "YES" << '\n';
    return 0;
  }
  vector<vector<long long> > dp(n + 1, vector<long long>(m, 0));
  for (long long i = n - 1; i >= 0; i--) {
    dp[i][v[i] % m] = 1;
    for (long long j = 0; j < m; j++) {
      if (dp[i + 1][j]) {
        dp[i][j] = 1;
        dp[i][(j + v[i] % m) % m] = 1;
      }
    }
    if (dp[i][0]) {
      cout << "YES" << '\n';
      return 0;
    }
  }
  cout << "NO";
  return 0;
}
