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>
unsigned GCD(unsigned u, unsigned v) {
  while (v != 0) {
    unsigned r = u % v;
    u = v;
    v = r;
  }
  return u;
}
int D[101000];
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &D[i]);
  }
  int x = GCD(D[0], D[1]), i;
  if (x != 1) {
    for (i = 2; i < N; i++) {
      x = GCD(x, D[i]);
      if (x == 1) break;
    }
    if (i == N) {
      printf("YES\n0\n");
      return 0;
    }
  }
  int cnt = 0;
  for (i = 0; i < N - 1; i++) {
    if (D[i] % 2 == 0) continue;
    if (D[i + 1] % 2 == 1) {
      cnt++;
      D[i + 1] = 2;
      continue;
    }
    cnt += 2;
    D[i + 1] = 2;
  }
  if (D[N - 1] % 2 == 1) {
    cnt += 2;
  }
  printf("YES\n%d\n", cnt);
  return 0;
}
