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;
long long n, m, arr[1000005], a[1005], mem[1001][1001];
bool res = 0;
bool solve(int idx, long long sum) {
  if (idx > n) return 0;
  if (!(sum % m) && sum != 0) return 1;
  long long &ret = mem[idx][sum % m];
  if (ret != -1) return ret;
  return ret = solve(idx + 1, sum) || solve(idx + 1, sum + arr[idx]);
}
int main() {
  memset(mem, -1, sizeof mem);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 0) return (cout << "YES" << endl, 0);
  }
  if (n > m) {
    for (int i = 0; i < n; i++) {
      if (!i) arr[i] %= m;
      (arr[i] += arr[i - 1]) %= m;
      a[arr[i]]++;
    }
    for (int i = 0; i <= m; i++)
      if (a[i] > 1) return (cout << "YES" << endl, 0);
    cout << "NO" << endl;
  } else {
    res = solve(0, 0);
    if (res)
      return (cout << "YES" << endl, 0);
    else
      cout << "NO" << endl;
  }
}
