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[1000007], x, y;
int dp[2000][2000];
bool solve(int idx, int sum) {
  if (sum && sum % y == 0) return 1;
  if (idx == x) return 0;
  int &ret = dp[idx][sum];
  if (ret != -1) return ret;
  return ret = solve(idx + 1, sum + arr[idx]) || solve(idx + 1, sum);
}
int main() {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  long long sum = 0;
  cin >> x >> y;
  for (int i = 0; i < x; i++) {
    cin >> arr[i];
    arr[i] %= y;
    if (!arr[i]) sum++;
  }
  memset(dp, -1, sizeof(dp));
  sort(arr, arr + x);
  if (solve(0, 0) || sum)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
