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>
#pragma GCC optimize("O3")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long N = 1e6 + 5;
long long a[N], n, m;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  if (n >= m) return cout << "YES\n", 0;
  vector<vector<long long> > dp(m, vector<long long>(2));
  dp[0][1] = 1;
  for (long long i = n; i >= 1; i--) {
    vector<vector<long long> > ndp(m, vector<long long>(2));
    for (long long j = 0; j < m; j++) {
      ndp[j][0] = (dp[j][0] | dp[(j + a[i]) % m][1]);
      ndp[j][1] = (dp[j][1] | dp[(j + a[i]) % m][1]);
    }
    dp = ndp;
  }
  cout << (dp[0][0] ? "YES" : "NO") << "\n";
  return 0;
}
