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 x = 0, f = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return f ? -x : x;
}
int gcd(int x, int y) {
  if (!x) return y;
  if (!y) return x;
  return gcd(y, x % y);
}
int a[100020], sum, ans;
int main(int argc, char const *argv[]) {
  int n = read();
  for (int i = 1; i <= n; i++) sum += (a[i] = read()) & 1;
  int gg = a[1];
  for (int i = 2; i <= n; i++) gg = gcd(gg, a[i]);
  if (gg > 1) return puts("YES\n0") & 0;
  for (int i = 1; i <= n - 1; i++)
    if ((a[i] & 1) && (a[i + 1] & 1)) {
      a[i] ^= 1;
      a[i + 1] ^= 1;
      sum -= 2, ans++;
    }
  printf("YES\n%d\n", ans + sum * 2);
  return 0;
}
