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) { return b == 0 ? a : GCD(b, a % b); }
int num[100005];
int main() {
  int n;
  while (cin >> n) {
    for (int j = 0; j < n; j++) cin >> num[j];
    int gcd = GCD(num[0], num[1]);
    for (int j = 2; j < n; j++) gcd = GCD(gcd, num[j]);
    if (gcd != 1) {
      cout << "YES" << endl;
      cout << "0" << endl;
      continue;
    }
    int ans = 0;
    bool odd_ok;
    if ((num[0] & 1) == 0)
      odd_ok = false;
    else
      odd_ok = true;
    for (int j = 1; j < n; j++) {
      if ((num[j] & 1) == 1) {
        if (odd_ok)
          ans++, odd_ok = false;
        else
          odd_ok = true;
      } else {
        if (odd_ok)
          ans += 2, odd_ok = false;
        else
          odd_ok = false;
      }
    }
    if (odd_ok) ans += 2;
    cout << "YES" << endl;
    cout << ans << endl;
  }
}
