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;
const int MAX = 1000 + 10;
const int LEN = 3;
bool dp[LEN][MAX];
int main() {
  memset(dp, sizeof(dp), 0);
  dp[0][0] = true;
  long long sum, target;
  cin >> sum >> target;
  for (int index = 1; index <= sum; index++) {
    long long a;
    cin >> a;
    for (long long i = target - 1; i >= 0; i--)
      if (dp[(index - 1) % 2][i]) {
        dp[index % 2][i] = true;
        dp[index % 2][(i + a) % target] = true;
        if ((i + a) % target == 0) {
          cout << "YES" << endl;
          return 0;
        }
      }
  }
  cout << "NO" << endl;
  return 0;
}
