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 M = 1E9 + 7;
inline long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
int f[200000][2], n;
long long a[200000], check;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  cin >> n;
  for (int i = 1; i <= n; cin >> a[i], i++)
    ;
  check = abs(a[1]);
  for (int i = 2; i <= n; check = GCD(abs(a[i]), check), i++)
    ;
  if (check > 1) {
    puts("YES\n0");
    return 0;
  }
  memset(f, 0x3f, sizeof(f)), f[1][a[1] & 1] = 0;
  for (int i = 2; i <= n; i++)
    if (a[i] & 1)
      f[i][1] = f[i - 1][0], f[i][0] = min(f[i - 1][1] + 1, f[i - 1][0] + 2);
    else
      f[i][1] = 0x3f3f3f3f, f[i][0] = min(f[i - 1][0], f[i - 1][1] + 2);
  printf("YES\n%d\n", f[n][0]);
  return 0;
}
