#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long quickpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % m;
    }
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int a[200005];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int i, j;
  cin >> a[1] >> a[2];
  long long d = gcd(a[1], a[2]);
  for (i = 3; i <= n; i++) {
    cin >> a[i];
    d = gcd(d, a[i]);
  }
  if (d > 1) {
    cout << "YES" << endl;
    cout << 0 << endl;
  } else {
    int cnt = 0;
    for (i = 1; i < n; i++) {
      if (a[i] % 2 && a[i + 1] % 2) {
        a[i] = 0;
        a[i + 1] = 0;
        cnt++;
      }
    }
    for (i = 1; i < n; i++) {
      if (a[i] % 2 && a[i + 1] % 2 == 0 || a[i + 1] % 2 && a[i] % 2 == 0) {
        a[i] = 0;
        a[i + 1] = 0;
        cnt += 2;
      }
    }
    cout << "YES" << endl;
    cout << cnt << endl;
  }
}

### Prompt

Your task is to create 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 int mod = 1e9 + 7;
long long quickpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % m;
    }
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int a[200005];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int i, j;
  cin >> a[1] >> a[2];
  long long d = gcd(a[1], a[2]);
  for (i = 3; i <= n; i++) {
    cin >> a[i];
    d = gcd(d, a[i]);
  }
  if (d > 1) {
    cout << "YES" << endl;
    cout << 0 << endl;
  } else {
    int cnt = 0;
    for (i = 1; i < n; i++) {
      if (a[i] % 2 && a[i + 1] % 2) {
        a[i] = 0;
        a[i + 1] = 0;
        cnt++;
      }
    }
    for (i = 1; i < n; i++) {
      if (a[i] % 2 && a[i + 1] % 2 == 0 || a[i + 1] % 2 && a[i] % 2 == 0) {
        a[i] = 0;
        a[i + 1] = 0;
        cnt += 2;
      }
    }
    cout << "YES" << endl;
    cout << cnt << endl;
  }
}

```