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;
bool d[1001];
int my_mod(int a, int m) {
  if (a % m == 0)
    return m;
  else
    return a % m;
}
int main() {
  int n, m, a[1000010];
  scanf("%d%d", &n, &m);
  vector<int> v;
  bool ok = false;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] %= m;
    if (a[i] == 0) ok = true;
  }
  if (ok || n >= m) {
    cout << "YES" << endl;
  } else {
    d[0] = true;
    v.push_back(0);
    for (int i = 0; i < n; i++) {
      if (ok) break;
      int v_size = v.size();
      for (int j = 0; j < v_size; j++) {
        int my_mod_n = my_mod(v[j] + a[i], m);
        if (my_mod_n == m) {
          ok = true;
          break;
        }
        if (!d[my_mod_n]) {
          d[my_mod_n] = true;
          v.push_back(my_mod_n);
        }
      }
    }
    if (ok) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
  return 0;
}
