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>
int N, a[100000], G, O;
int gcd(int x, int y) {
  int r;
  while (y) {
    r = x % y;
    x = y;
    y = r;
  }
  return x;
}
inline int F(int x) { return (x >> 1) + ((x & 1) << 1); }
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", a + i);
  G = a[0];
  for (int i = 1; i < N; i++) G = gcd(G, a[i]);
  if (G > 1) {
    puts("YES\n0");
    return 0;
  }
  for (int i = 0; i < N; i++)
    if (a[i] & 1) {
      int j = i;
      while (j < N - 1 && (a[j + 1] & 1)) j++;
      O += F(j - i + 1);
      i = j;
    }
  printf("YES\n%d\n", O);
  return 0;
}
