#include <bits/stdc++.h>
using namespace std;
const int MX = 1234567;
const double PI = acos(-1.0), EPS = 1e-9;
long long N, arr[MX];
long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
int main() {
  cin >> N;
  long long g = 0;
  for (int i = 0; i < N; i++) cin >> arr[i];
  for (int i = 0; i < N; i++) {
    g = gcd(g, arr[i]);
  }
  if (g != 1) {
    cout << "YES\n";
    cout << 0 << "\n";
    return 0;
  }
  int oper = 0;
  for (int i = 0; i < N; i++) {
    if (arr[i] % 2 && arr[i + 1] % 2) {
      i++;
      oper++;
    } else if (arr[i] % 2 && arr[i + 1] % 2 == 0) {
      i++;
      oper += 2;
    }
  }
  cout << "YES\n";
  cout << oper << "\n";
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
using namespace std;
const int MX = 1234567;
const double PI = acos(-1.0), EPS = 1e-9;
long long N, arr[MX];
long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
int main() {
  cin >> N;
  long long g = 0;
  for (int i = 0; i < N; i++) cin >> arr[i];
  for (int i = 0; i < N; i++) {
    g = gcd(g, arr[i]);
  }
  if (g != 1) {
    cout << "YES\n";
    cout << 0 << "\n";
    return 0;
  }
  int oper = 0;
  for (int i = 0; i < N; i++) {
    if (arr[i] % 2 && arr[i + 1] % 2) {
      i++;
      oper++;
    } else if (arr[i] % 2 && arr[i + 1] % 2 == 0) {
      i++;
      oper += 2;
    }
  }
  cout << "YES\n";
  cout << oper << "\n";
  return 0;
}

```