#include <bits/stdc++.h>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n;
  std::cin >> n;
  std::vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> v[i];
  }
  int g = v[0];
  for (int j = 0; j < v.size(); ++j) {
    g = gcd(g, v[j]);
    if (g == 1) {
      break;
    }
  }
  if (g > 1) {
    std::cout << "YES" << std::endl << 0 << std::endl;
    return 0;
  }
  int count = 0;
  int num_odd = 0;
  for (int k = 0; k < v.size(); ++k) {
    if (v[k] % 2) {
      num_odd++;
    } else {
      count += num_odd / 2 + num_odd % 2 * 2;
      num_odd = 0;
    }
  }
  count += num_odd / 2 + num_odd % 2 * 2;
  std::cout << "YES" << std::endl << count << std::endl;
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n;
  std::cin >> n;
  std::vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    std::cin >> v[i];
  }
  int g = v[0];
  for (int j = 0; j < v.size(); ++j) {
    g = gcd(g, v[j]);
    if (g == 1) {
      break;
    }
  }
  if (g > 1) {
    std::cout << "YES" << std::endl << 0 << std::endl;
    return 0;
  }
  int count = 0;
  int num_odd = 0;
  for (int k = 0; k < v.size(); ++k) {
    if (v[k] % 2) {
      num_odd++;
    } else {
      count += num_odd / 2 + num_odd % 2 * 2;
      num_odd = 0;
    }
  }
  count += num_odd / 2 + num_odd % 2 * 2;
  std::cout << "YES" << std::endl << count << std::endl;
  return 0;
}

```