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;
bool solve(const int n, const int m) {
  vector<int> a(n);
  for (int &x : a) {
    cin >> x;
  }
  if (n > m) {
    return true;
  }
  vector<bool> prev_dp(m, false);
  for (const int x : a) {
    auto dp = prev_dp;
    dp[x % m] = true;
    for (int i = 0; i < prev_dp.size(); ++i) {
      if (prev_dp[i]) {
        dp[(i + x) % m] = true;
      }
    }
    if (dp[0]) {
      return true;
    }
    prev_dp = dp;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  while (cin >> n >> m) {
    if (solve(n, m)) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}
