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 N = 1e6 + 5;
long long n, m;
long long arr[N];
bool dp[1005][1005], vis[1005][1005];
bool solve(int num, int sum, bool take) {
  if (num == n) {
    return (sum % m == 0 && take);
  }
  sum %= m;
  bool &ret = dp[num][sum];
  if (vis[num][sum]) return ret;
  vis[num][sum] = 1;
  bool c1 = solve(num + 1, sum + arr[num], 1);
  bool c2 = solve(num + 1, sum, take);
  return ret = (c1 || c2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i] %= m;
  }
  if (n > m) return cout << "YES" << endl, 0;
  memset(dp, 0, sizeof dp);
  memset(vis, 0, sizeof vis);
  if (solve(0, 0, 0))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
