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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long INF = 1e18 + 1LL;
const double Pi = acos(-1.0);
const int N = 1e6 + 10, M = 1e3 + 20, mod = 1e9 + 7, inf = 2e9;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int a[N], b[N], n;
set<long long> s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  long long G = gcd(a[1], a[2]);
  for (int i = 2; i <= n; ++i) {
    G = gcd(G, a[i]);
  }
  if (G != 1) {
    puts("YES");
    cout << 0 << endl;
    return 0;
  }
  puts("YES");
  int ans = 0;
  for (int i = 1; i <= n; i += 1) {
    if (a[i] % 2 == 1 && a[i + 1] % 2 == 0)
      ans += 2, a[i] = 0;
    else if (a[i] % 2 == 1 && a[i + 1] % 2 == 1)
      ans++, a[i + 1] = 0;
  }
  cout << ans << endl;
  return 0;
}
