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 = 1e3 + 5;
int n, m, frq[N], dp[N][N][2];
int solve(int num = 0, int sum = 0, bool go = 0) {
  if (!sum && go) {
    cout << "YES";
    exit(0);
  }
  if (num == m) return 0;
  int &ret = dp[num][sum][go];
  if (~ret) return ret;
  int ans = 0;
  for (int i = 0; i <= frq[num]; ++i)
    ans |= solve(num + 1, (sum + num * i) % m, i | go);
  return ret = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  memset(dp, -1, sizeof dp);
  while (n--) {
    int x;
    cin >> x;
    ++frq[x % m];
  }
  solve();
  cout << "NO";
  return 0;
}
