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 ll = long long;
int n, m;
ll a;
int main() {
  std::cin >> n >> m;
  std::set<int> vals;
  std::set<int> extras;
  bool can_do = false;
  for (int i = 0; i < n; i++) {
    std::cin >> a;
    a %= m;
    for (const auto& i : vals) extras.insert((i + a) % m);
    for (const auto& val : extras) vals.insert(val);
    extras.clear();
    vals.insert(a);
    can_do = vals.find(0) != vals.end();
    if (can_do) break;
  }
  std::cout << (can_do ? "YES" : "NO") << std::endl;
}
