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. 
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  while (a != 0) {
    int c = a;
    a = b % a;
    b = c;
  }
  return b;
}
vector<int> adj[3000];
vector<int> getAllDivisors(int num) {
  vector<int> divs;
  int square_root = (int)sqrt(num) + 1;
  for (int i = 1; i < square_root; i++) {
    if (num % i == 0) {
      if (i * i != num) {
        divs.push_back(i);
      }
      divs.push_back(num / i);
    }
  }
  return divs;
}
class ChangingChange {
 public:
};
long long int mod_pow(long x, int y) {
  long long int r = 1;
  while (y > 0) {
    if (y % 2 == 1) {
      r = (r * x) % (1000000007);
    }
    x = (x * x) % (1000000007);
    y /= 2;
  }
  return r;
}
int main() {
  ChangingChange temp;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  vector<int> arr;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    arr.push_back(temp);
  }
  int moves = 0;
  int GCD = arr[0];
  for (int i = 1; i < n; i++) {
    GCD = gcd(GCD, arr[i]);
  }
  if (GCD > 1) {
    cout << "YES" << '\n' << moves;
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if ((arr[i] % 2 == 1) && (arr[i + 1] % 2 == 1)) {
      arr[i] = 2, arr[i + 1] = 2;
      moves++;
    }
  }
  for (int i = 0; i < n; i++) {
    if (arr[i] % 2) {
      moves += 2;
    }
  }
  cout << "YES" << '\n' << moves;
  return 0;
}
