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;
vector<long long> v;
long long b[10000];
void calc(long long x) {
  if (!b[x]) {
    b[x] = 1;
    v.push_back(x);
  }
}
int main() {
  long long n, k, f = 0, x;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    long long s = v.size();
    for (long long j = 0; j < s; j++) calc((v[j] + x) % k);
    calc(x % k);
    if (b[0]) {
      f = 1;
      break;
    }
  }
  if (f == 1)
    cout << "YES";
  else
    cout << "NO";
}
