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 arr[1000000] = {}, a[20111000] = {};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int x, ans = 0, y;
  int i, j, z, c = 0;
  cin >> x >> y;
  for (i = 0; i < x; i++) {
    cin >> z;
    arr[i] = z % y;
    if (arr[i] == 0) c = 1;
  }
  if (c) {
    cout << "YES";
    return 0;
  }
  a[0] = 1;
  for (i = x - 1; i >= 0; i--) {
    for (j = 1000111; j >= 0; j--) {
      if (a[j] == 1 && (j + arr[i]) % y == 0) {
        cout << "YES";
        return 0;
      }
      if (a[j] == 1) {
        a[(j + arr[i])] = a[j];
      }
    }
  }
  if (ans)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
