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;
const long long int MOD = 1000000007;
int N;
list<int> P;
int ans = INT_MAX;
void Search(vector<long long int> v, int p) {
  int cnt = 0;
  for (int i = 0; i < (signed)v.size() - 1; i++) {
    if ((v[i] + p * MOD) % p) {
      long long int box[2] = {v[i] - v[i + 1], v[i] + v[i + 1]};
      v[i] = box[0];
      v[i + 1] = box[1];
      if ((v[i] + p * MOD) % p) return;
      cnt++;
    }
  }
  ans = min(ans, cnt);
}
int main() {
  cin >> N;
  vector<long long int> num(N + 1);
  for (int i = 0; i < N; i++) cin >> num[i];
  P.push_back(2);
  for (int i = 3; i <= sqrt(N); i += 2) {
    bool flag = true;
    for (auto j : P) {
      if (!(i % j)) {
        flag = false;
        break;
      }
    }
    if (flag) P.push_back(i);
  }
  for (int i = 0; i < N; i++) {
    int box = num[i];
    for (auto j : P) {
      while (box % j == 0) {
        box /= j;
      }
    }
    if (box != 1) P.push_back(box);
  }
  for (auto i : P) {
    bool flag = true;
    for (int j = 0; j < N; j++) {
      if (num[j] % i) {
        flag = false;
        break;
      }
    }
    if (flag) {
      cout << "YES\n";
      cout << "0\n";
      return 0;
    }
  }
  for (int i = 0; i < N; i++) {
    num[i] %= 2;
  }
  ans = 0;
  for (int i = 0; i < N; i++) {
    if (num[i] && num[i + 1]) {
      ans++;
      i++;
    } else if (num[i]) {
      ans += 2;
      i++;
    }
  }
  cout << "YES\n";
  cout << ans << endl;
}
