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>
const int mod = 1000000007;
using namespace std;
map<vector<long long>, int> mp;
long long flag = 0, n, m;
void findAns(vector<long long> &ar, long long index, long long sum) {
  if (flag == 1 || mp[{index, sum}] == 1 || index == n) return;
  mp[{index, sum}] = 1;
  long long sum1 = (sum + ar[index]) % m;
  if (sum1 % m == 0) {
    flag = 1;
    return;
  }
  findAns(ar, index + 1, sum1);
  findAns(ar, index + 1, sum);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  vector<long long> ar(n);
  for (long long i = 0; i < n; i++) {
    cin >> ar[i];
    ar[i] %= m;
    if (ar[i] == 0) flag = 1;
  }
  findAns(ar, 0, 0);
  if (flag)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
