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 inf = 0x3f3f3f3f;
const unsigned long long INF = 0x3f3f3f3f3f3f3f3fLL;
const long long maxn = 1e6 + 20;
long long n, m;
set<long long> has;
long long a[maxn];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long t;
    cin >> t;
    a[i] = t % m;
    if (a[i] == 0) {
      cout << "YES";
      return 0;
    }
  }
  for (long long i = 0; i < n; i++) {
    vector<long long> vec;
    for (long long t : has) {
      long long num = (a[i] + t) % m;
      if (num == 0) {
        cout << "YES";
        return 0;
      } else {
        vec.push_back(num);
      }
    }
    has.insert(a[i]);
    has.insert(begin(vec), end(vec));
  }
  cout << "NO";
  return 0;
}
