#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> v;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  scanf("%d", &n);
  long long g = 0;
  for (int i = int(0); i < (int)n; ++i) {
    int a;
    scanf("%d", &a);
    v.push_back(a);
    g = gcd(g, a);
  }
  if (g > 1) {
    printf("YES\n%d\n", 0);
    return 0;
  }
  int ans = 0;
  for (int i = int(0); i < (int)n - 1; ++i) {
    int a, b;
    while (v[i] & 1) {
      a = v[i], b = v[i + 1];
      v[i] = a + b;
      v[i + 1] = a - b;
      ans++;
    }
  }
  int a, b;
  while (v[n - 1] & 1) {
    a = v[n - 2], b = v[n - 1];
    v[n - 2] = a + b;
    v[n - 1] = a - b;
    ans++;
  }
  printf("YES\n%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int n;
vector<int> v;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  scanf("%d", &n);
  long long g = 0;
  for (int i = int(0); i < (int)n; ++i) {
    int a;
    scanf("%d", &a);
    v.push_back(a);
    g = gcd(g, a);
  }
  if (g > 1) {
    printf("YES\n%d\n", 0);
    return 0;
  }
  int ans = 0;
  for (int i = int(0); i < (int)n - 1; ++i) {
    int a, b;
    while (v[i] & 1) {
      a = v[i], b = v[i + 1];
      v[i] = a + b;
      v[i + 1] = a - b;
      ans++;
    }
  }
  int a, b;
  while (v[n - 1] & 1) {
    a = v[n - 2], b = v[n - 1];
    v[n - 2] = a + b;
    v[n - 1] = a - b;
    ans++;
  }
  printf("YES\n%d\n", ans);
  return 0;
}

```