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 an[1010000];
bool visit[2][1100];
int main() {
  int z;
  int n, m;
  int i, j, k;
  cin >> n >> m;
  for (i = 1; i <= n; i++) scanf("%d", &an[i]);
  if (n >= m)
    cout << "YES" << endl;
  else {
    memset(visit, 0, sizeof(visit));
    visit[0][0] = 1;
    for (i = 1; i <= n; i++) {
      int md = an[i] % m;
      for (j = 0; j < m; j++) {
        if (visit[i - 1 & 1][j]) {
          visit[i & 1][(j + md) % m] = 1;
          if ((j + md) % m == 0) {
            cout << "YES" << endl;
            return 0;
          }
        }
      }
      for (j = 0; j < m; j++) {
        if (visit[i - 1 & 1][j]) visit[i & 1][j] = 1;
      }
    }
    cout << "NO" << endl;
  }
  return 0;
}
