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;
inline void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int prm(long long int a, long long int b, long long int m) {
  long long int ans = ((a % m) * (b % m)) % m;
  return ans;
}
long long int smm(long long int a, long long int b, long long int m) {
  long long int ans = (a + b) % m;
  return ans;
}
long long int sbm(long long int a, long long int b, long long int m) {
  long long int ans = (a - b + m) % m;
  return ans;
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  long long int a[n + 1];
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = a[i] % m;
  }
  unordered_map<long long int, long long int> pp;
  for (long long int i = 1; i <= n; i++) {
    if (i == 1) {
      pp[a[i]] = 1;
    } else {
      long long int req = m - a[i] % m;
      if (pp.find(req) != pp.end()) {
        cout << "YES";
        return;
      } else {
        vector<long long int> ins;
        if (pp.find(a[i]) == pp.end()) {
          ins.push_back(a[i]);
        }
        for (auto it : pp) {
          long long int tt = it.first;
          long long int sum = (tt + a[i]) % m;
          if (pp.find(sum) == pp.end()) {
            ins.push_back(sum);
          }
        }
        for (long long int it : ins) pp[it] = 1;
      }
    }
  }
  if (pp[0] == 1) {
    cout << "YES";
    return;
  } else {
    cout << "NO";
    return;
  }
}
int main() {
  fast();
  solve();
}
