Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>. 

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2
1 1


Output

YES
1


Input

3
6 2 4


Output

YES
0


Input

2
1 3


Output

YES
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with <image>.

In the second example the gcd of the sequence is already greater than 1. 
#include <bits/stdc++.h>
const double dinf = std::numeric_limits<double>::max();
struct pairhash {
  template <typename T, typename U>
  std::size_t operator()(const std::pair<T, U>& x) const {
    return std::hash<T>()(x.first) ^ std::hash<U>()(x.second);
  }
};
long long compute_gcd(long long a, long long b) {
  if (b == 0) return a;
  return compute_gcd(b, a % b);
}
unsigned long long compute_lcm(unsigned long long a, unsigned long long b) {
  unsigned long long gcd = compute_gcd(a, b);
  return a / gcd * b;
}
int mod_mul(int a, int b, int mod) { return ((long long)a * b) % mod; }
int mod_add(int a, int b, int mod) { return ((long long)a + b) % mod; }
long long choose(int n, int k) {
  if (k == 0 || n == 0) return 1;
  long long result = n * choose(n - 1, k - 1) / k;
  return result;
}
using namespace std;
bool initialCheck(vector<int> const& numbers) {
  int gcd = numbers[0];
  for (int i = 1; i < numbers.size(); ++i) {
    gcd = compute_gcd(gcd, numbers[i]);
    if (gcd == 1) return false;
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  vector<int> numbers(n);
  for (int i = 0; i < n; ++i) cin >> numbers[i];
  if (initialCheck(numbers)) {
    cout << "YES\n" << 0;
    return 0;
  }
  numbers.push_back(2);
  int moves = 0;
  for (int i = 0; i < numbers.size() - 1;) {
    auto& current = numbers[i];
    auto& next = numbers[i + 1];
    if (current % 2 == 0) {
      ++i;
      continue;
    }
    int prevCurrent = current;
    int prevNext = next;
    current = prevCurrent - prevNext;
    next = prevCurrent + prevNext;
    ++moves;
  }
  cout << "YES\n" << moves;
  return 0;
}
