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 a[1111];
int n, p;
int f[1111][1111];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> p;
  if (n >= p) {
    cout << "YES";
    exit(0);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = a[i] % p;
    if (a[i] == 0) {
      cout << "YES";
      exit(0);
    }
  }
  for (int i = 1; i <= n; i++) {
    f[i][a[i]] = 1;
    for (int j = 0; j <= p; j++) {
      if (f[i][j] == 1) continue;
      if (j - a[i] >= 0)
        f[i][j] = (f[i - 1][j] | f[i - 1][j - a[i]]);
      else
        f[i][j] = f[i - 1][j] | f[i - 1][j - a[i] + p];
    }
  }
  if (f[n][p] == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
