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() {
  std::ios::sync_with_stdio(false);
  int n, m, arr[1000001], sum = 0, p = 0, ct = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i] = arr[i] % m;
    if (arr[i] == 0) ct++;
    sum += arr[i] % m;
  }
  bool dp[1000010] = {0};
  if (n > m)
    cout << "YES";
  else {
    dp[0] = true;
    for (int i = 0; i < n; i++)
      for (int j = sum; j >= arr[i]; j--) dp[j] = dp[j] | dp[j - arr[i]];
    for (int i = 0; i <= sum; i++)
      if (dp[i] == true && i >= m && i % m == 0 || (i == 0 && ct != 0)) p = 1;
    if (p)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}
