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 dTb(long long n, vector<long long> &v) {
  for (long long i = 0; i < 20; i++) {
    v[i] = n & 1;
    n = n >> 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long i, j, k, m, n, t, q, r, c, x;
  cin >> n >> m;
  if (n > m) {
    cout << "YES";
    return 0;
  }
  vector<long long> v(n);
  vector<vector<long long> > dp(n, vector<long long>(m, 0));
  for (long long(i) = (0); (i) < (n); ++(i)) {
    cin >> v[i];
    v[i] = v[i] % m;
    if (v[i] == 0) {
      cout << "YES";
      return 0;
    }
  }
  dp[0][v[0]] = 1;
  for (long long(i) = (1); (i) < (n); ++(i)) {
    for (long long(j) = (0); (j) < (m); ++(j)) {
      dp[i][j] = max(dp[i - 1][j], dp[i][j]);
      if (dp[i - 1][j]) {
        dp[i][(j + v[i]) % m] = 1;
      }
      dp[i][v[i]] = 1;
    }
  }
  if (dp[n - 1][0]) {
    cout << "YES";
    return 0;
  }
  cout << "NO";
}
