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>
int dx[] = {0, -1, 0, 1, -1, 1, -1, 1};
int dy[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int MOD = 1e9 + 7, N = 1e6 + 5;
const double pi = acos(-1);
using namespace std;
int n, m, arr[N];
int dp[1005][1005];
bool solve(int id, long long sum, int taken) {
  if (!sum and taken) return true;
  if (~dp[id][sum]) return dp[id][sum];
  if (id == n) return false;
  bool ans = solve(id + 1, sum, taken);
  ans |= solve(id + 1, (sum + arr[id]) % m, taken + 1);
  return dp[id][sum] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset(dp, -1, sizeof dp);
  cin >> n >> m;
  bool zero = false;
  if (n > m) return cout << "YES", 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    zero |= !arr[i];
  }
  (solve(0, 0, 0) or zero) ? cout << "YES" : cout << "NO";
  return 0;
}
