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;
const long long int N = 5e5 + 7;
const long long int inf = 1e10 + 7;
const long long int mod = 1e9 + 7;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  gcd(b % a, a);
}
int main() {
  long long int n, i, j, k = 0, ok = 1, ans = 0, ret;
  cin >> n;
  long long int a[n + 1];
  for (i = 0; i < n; i++) scanf("%lld", &a[i]);
  ret = a[0];
  for (i = 1; i < n; i++) {
    ret = gcd(ret, a[i]);
  }
  if (ret != 1) {
    printf("YES\n0\n");
    return 0;
  }
  ok = 1;
  for (i = 0; i < n - 1; i++) {
    long long int x = a[i];
    long long int y = a[i + 1];
    if (a[i] % 2 == 0)
      continue;
    else {
      if (a[i + 1] & 1) {
        ans++;
        a[i] = x - y;
        a[i + 1] = x + y;
      } else {
        ans += 2;
        a[i] = 2 * y;
        a[i + 1] = 2 * x;
      }
    }
  }
  if (a[n - 1] % 2 != 0) {
    a[n - 1] = 2;
    ans += 2;
  }
  for (i = 0; i < n; i++) {
    if (a[i] % 2 != 0) ok = 0;
  }
  if (!ok)
    printf("NO\n");
  else {
    printf("YES\n%lld\n", ans);
  }
  return 0;
}
