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 N = 1e6 + 5;
const int M = 1e3 + 5;
int A[N];
bool dp[2][M];
int n, m;
void solve() {
  cin >> n >> m;
  if (n > m)
    cout << "YES"
         << "\n",
        exit(0);
  ;
  for (int i = 1; i <= n; i++) cin >> A[i];
  int i = 0;
  dp[i][0] = true;
  for (int k = 1; k <= n; k++) {
    for (int r = 0; r < m; r++) {
      int t = (r + A[k]) % m;
      dp[i ^ 1][t] |= dp[i][r];
      if (dp[i ^ 1][t] && t == 0)
        cout << "YES"
             << "\n",
            exit(0);
      ;
    }
    for (int r = 0; r < m; r++) dp[i ^ 1][r] |= dp[i][r];
    fill(&dp[i][0], &dp[i][0] + m, false);
    i ^= 1;
  }
  cout << "NO"
       << "\n",
      exit(0);
  ;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
