#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e5 + 10;
int gcd(int n, int m) { return m == 0 ? n : gcd(m, n % m); }
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto &x : v) cin >> x;
  v.push_back(v.front() * 2);
  int g = v.front();
  for (auto &x : v) g = gcd(x, g);
  cout << "YES" << endl;
  if (g != 1) {
    cout << 0 << endl;
    return 0;
  }
  int ans = 0;
  int cnt = 0;
  for (auto &x : v) {
    if (x & 1)
      cnt++;
    else
      ans += (cnt / 2) + (cnt & 1) * 2, cnt = 0;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e5 + 10;
int gcd(int n, int m) { return m == 0 ? n : gcd(m, n % m); }
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (auto &x : v) cin >> x;
  v.push_back(v.front() * 2);
  int g = v.front();
  for (auto &x : v) g = gcd(x, g);
  cout << "YES" << endl;
  if (g != 1) {
    cout << 0 << endl;
    return 0;
  }
  int ans = 0;
  int cnt = 0;
  for (auto &x : v) {
    if (x & 1)
      cnt++;
    else
      ans += (cnt / 2) + (cnt & 1) * 2, cnt = 0;
  }
  cout << ans << endl;
  return 0;
}

```