#include <bits/stdc++.h>
using namespace std;
int gcd(int x, int y) { return x % y == 0 ? y : gcd(y, x % y); }
int a[100050];
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  int i, j;
  cin >> a[1];
  int G = a[1];
  for (i = 2; i <= N; i++) {
    cin >> a[i];
    G = gcd(a[i], G);
  }
  if (G > 1) {
    cout << "YES" << endl << 0 << endl;
    exit(0);
  }
  int Count = 0;
  a[1] = a[1] % 2;
  for (i = 2; i <= N; i++) {
    a[i] = a[i] % 2;
    if (a[i] && a[i - 1]) {
      a[i] = 0;
      a[i - 1] = 0;
      Count++;
    }
  }
  for (i = 2; i <= N; i++) {
    if (a[i] + a[i - 1] == 1) {
      a[i] = 0;
      a[i - 1] = 0;
      Count += 2;
    }
  }
  cout << "YES" << endl << Count << endl;
  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;
int gcd(int x, int y) { return x % y == 0 ? y : gcd(y, x % y); }
int a[100050];
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  int i, j;
  cin >> a[1];
  int G = a[1];
  for (i = 2; i <= N; i++) {
    cin >> a[i];
    G = gcd(a[i], G);
  }
  if (G > 1) {
    cout << "YES" << endl << 0 << endl;
    exit(0);
  }
  int Count = 0;
  a[1] = a[1] % 2;
  for (i = 2; i <= N; i++) {
    a[i] = a[i] % 2;
    if (a[i] && a[i - 1]) {
      a[i] = 0;
      a[i - 1] = 0;
      Count++;
    }
  }
  for (i = 2; i <= N; i++) {
    if (a[i] + a[i - 1] == 1) {
      a[i] = 0;
      a[i - 1] = 0;
      Count += 2;
    }
  }
  cout << "YES" << endl << Count << endl;
  return 0;
}

```