#include <bits/stdc++.h>
using namespace std;
const int maxm = 100005;
int a[maxm], b[maxm];
int gcd(int x, int y);
int main() {
  int n, i, j, k, sum, ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  if (b[n] == 0)
    ans = 5;
  else {
    ans = b[n];
    for (i = n - 1; i >= 1; i--) {
      if (b[i] == 0) continue;
      ans = gcd(ans, b[i]);
    }
  }
  if (ans > 1) {
    printf("YES\n0\n");
    return 0;
  }
  ans = 0;
  for (i = 1; i < n; i++) {
    if (a[i] % 2 == 0) continue;
    if (a[i + 1] % 2 != 0) {
      ans++;
      a[i + 1] = 2;
    } else {
      ans += 2;
      a[i + 1] = 2;
    }
  }
  if (a[n] % 2 != 0) ans += 2;
  printf("YES\n%d\n", ans);
  return 0;
}
int gcd(int x, int y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}

### 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 maxm = 100005;
int a[maxm], b[maxm];
int gcd(int x, int y);
int main() {
  int n, i, j, k, sum, ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  if (b[n] == 0)
    ans = 5;
  else {
    ans = b[n];
    for (i = n - 1; i >= 1; i--) {
      if (b[i] == 0) continue;
      ans = gcd(ans, b[i]);
    }
  }
  if (ans > 1) {
    printf("YES\n0\n");
    return 0;
  }
  ans = 0;
  for (i = 1; i < n; i++) {
    if (a[i] % 2 == 0) continue;
    if (a[i + 1] % 2 != 0) {
      ans++;
      a[i + 1] = 2;
    } else {
      ans += 2;
      a[i + 1] = 2;
    }
  }
  if (a[n] % 2 != 0) ans += 2;
  printf("YES\n%d\n", ans);
  return 0;
}
int gcd(int x, int y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}

```