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;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, a[100100], b, ans;
int gcd(int v1, int v2) {
  if (v1 == 0)
    return v2;
  else
    return gcd(v2 % v1, v1);
}
int main() {
  puts("YES");
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (i == 2) {
      b = gcd(a[1], a[2]);
    } else if (i > 2 && b > 1) {
      b = gcd(b, a[i]);
    }
  }
  if (b > 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] % 2) {
      if (a[i + 1] % 2) {
        ans++;
        a[i + 1] = 2;
      } else {
        ans += 2;
      }
    }
  }
  printf("%d", ans);
  return 0;
}
