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 x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
const int SIZE = 1e5 + 7;
int GCD, n, a[SIZE], ans;
void get_GCD() {
  for (int(i) = 0; (i) < (int)(n); (i)++) GCD = gcd(GCD, a[i]);
  if (GCD > 1) {
    cout << "YES" << endl;
    cout << "0" << endl;
    exit(0);
  }
  return;
}
void into(int x) {
  while (abs(a[x]) % 2 != 0) {
    ans++;
    int t1 = a[x] - a[x + 1];
    int t2 = a[x] + a[x + 1];
    a[x] = t1;
    a[x + 1] = t2;
  }
}
int main() {
  cin >> n;
  for (int(i) = 0; (i) < (int)(n); (i)++) cin >> a[i];
  get_GCD();
  for (int(i) = 0; (i) < (int)(n - 1); (i)++) into(i);
  while (abs(a[n - 1]) % 2 != 0) {
    ans++;
    int t1 = a[n - 2] - a[n - 1];
    int t2 = a[n - 2] + a[n - 1];
    a[n - 2] = t1;
    a[n - 1] = t2;
  }
  if (ans) {
    cout << "YES" << endl;
    cout << ans << endl;
  } else {
    cout << "NO" << endl;
    cout << "-1" << endl;
  }
  return 0;
}
