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 long long power(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long p = power(a, n / 2, m);
  p = (p * p) % m;
  if (n % 2)
    return (p * a) % m;
  else
    return p;
}
void show() {
  cout << "YES";
  exit(0);
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, m;
  cin >> n >> m;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  for (long long int i = 0; i < n; i++) {
    arr[i] = arr[i] % m;
    if (arr[i] == 0) {
      return cout << "YES", 0;
    }
  }
  if (n >= m) return cout << "YES", 0;
  set<long long int> d;
  for (long long int i = 0; i < n; i++) {
    set<long long int> tmp = d;
    for (auto it : tmp) {
      d.insert((it + arr[i]) % m);
    }
    d.insert(arr[i]);
    if (d.find(0) != d.end()) show();
  }
  cout << "NO";
}
