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 int MOD = 1e9 + 7;
const int di[] = {-1, 0, 1, 0, 1, -1, 1, -1};
const int dj[] = {0, 1, 0, -1, 1, -1, -1, 1};
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<int> num(n);
  int tg = 0;
  for (int i = 0; i < n; i++) cin >> num[i], tg = gcd(tg, num[i]);
  if (tg > 1) {
    cout << "YES" << endl << 0 << endl;
    return 0;
  }
  int a, b, r = 0;
  for (int i = 0; i < n - 1; i++) {
    if (abs(num[i]) % 2 == 1 && abs(num[i + 1]) % 2 == 1) {
      while (num[i] % 2 == 1 || num[i + 1] % 2 == 1) {
        a = num[i], b = num[i + 1];
        num[i] = a - b, num[i + 1] = a + b;
        r++;
      }
    }
  }
  for (int i = 0; i < n - 1; i++) {
    while (abs(num[i]) % 2 == 1) {
      a = num[i], b = num[i + 1];
      num[i] = a - b, num[i + 1] = a + b;
      r++;
    }
  }
  while (abs(num[n - 1]) % 2 == 1) {
    a = num[n - 2], b = num[n - 1];
    num[n - 2] = a - b, num[n - 1] = a + b;
    r++;
  }
  cout << "YES" << endl;
  cout << r << endl;
}
