#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, m;
int a[MAXN], b[MAXN];
int ans;
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  if (x % y != 0) {
    return gcd(y, x % y);
  }
  return y;
}
int main(int argc, const char* argv[]) {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(a[i]));
    b[i] = a[i] % 2;
  }
  b[n + 1] = 0;
  int tgcd = abs(gcd(a[1], a[2]));
  for (int i = 3; i <= n; i++) {
    tgcd = gcd(tgcd, a[i]);
  }
  cout << "YES" << endl;
  if (tgcd > 1) {
    cout << "0" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] == 1) {
      if (b[i + 1] == 1) {
        ans++;
        b[i] = 0;
        b[i + 1] = 0;
      } else {
        ans += 2;
        b[i] = 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 MAXN = 1e5 + 10;
int n, m;
int a[MAXN], b[MAXN];
int ans;
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  if (x % y != 0) {
    return gcd(y, x % y);
  }
  return y;
}
int main(int argc, const char* argv[]) {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(a[i]));
    b[i] = a[i] % 2;
  }
  b[n + 1] = 0;
  int tgcd = abs(gcd(a[1], a[2]));
  for (int i = 3; i <= n; i++) {
    tgcd = gcd(tgcd, a[i]);
  }
  cout << "YES" << endl;
  if (tgcd > 1) {
    cout << "0" << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (b[i] == 1) {
      if (b[i + 1] == 1) {
        ans++;
        b[i] = 0;
        b[i + 1] = 0;
      } else {
        ans += 2;
        b[i] = 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```