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;
long long getGCD(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<long long> vect(n);
  for (int i = 0; i < n; i++) cin >> vect[i];
  long long gcd = getGCD(vect[0], vect[1]);
  for (int i = 2; i < vect.size(); i++) {
    gcd = getGCD(gcd, vect[i]);
  }
  if (gcd != 1) {
    cout << "YES\n" << 0;
    return 0;
  }
  long long steps = 0;
  for (int i = 0; i < vect.size(); i++) {
    if (vect[i] % 2 != 0) {
      if (i < vect.size() - 1 && abs(vect[i + 1]) % 2 != 0) {
        steps++;
        long long temp = vect[i];
        vect[i] -= vect[i + 1];
        vect[i] = abs(vect[i]);
        vect[i + 1] = temp + vect[i + 1];
      }
    }
  }
  for (int i = 0; i < vect.size(); i++) {
    if (vect[i] % 2 != 0) {
      steps += 2;
    }
  }
  cout << "YES\n" << steps;
  return 0;
}
