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;
int main() {
  set<int> s;
  set<int>::iterator it;
  int a, b, c, d;
  scanf("%d %d", &a, &b);
  int arr[b + 10];
  for (int i = 0; i < a; i++) {
    set<int> se;
    cin >> arr[i];
    arr[i] %= b;
    if (arr[i] % b == 0) {
      printf("YES\n");
      return 0;
    }
    for (it = s.begin(); it != s.end(); it++) {
      int y = (*it + arr[i]) % b;
      if (y % b == 0) {
        printf("YES\n");
        return 0;
      }
      se.insert(y);
    }
    s.insert(arr[i]);
    for (it = se.begin(); it != se.end(); it++) {
      s.insert(*it);
    }
  }
  printf("NO\n");
}
