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>
const int N = 100050;
const int inf = 1e9 + 7;
int gcd(int x, int y) { return x ? gcd(y % x, x) : y; }
int min(int a, int b) { return a > b ? b : a; }
int dp[N][2], a[N];
int main() {
  int n, i, g = 0;
  scanf("%i", &n);
  for (i = 1; i <= n; i++) scanf("%i", &a[i]), g = gcd(g, a[i]), a[i] %= 2;
  if (g > 1) return printf("YES\n0\n"), 0;
  for (i = 0; i < N; i++) dp[i][0] = dp[i][1] = inf;
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    if (a[i]) {
      dp[i][0] = dp[i - 1][1] + 1;
      dp[i][0] = min(dp[i][0], dp[i - 1][0] + 2);
      dp[i][1] = dp[i - 1][0];
    } else {
      dp[i][0] = dp[i - 1][0];
      dp[i][0] = min(dp[i][0], dp[i - 1][1] + 2);
    }
  }
  printf("YES\n");
  printf("%i\n", dp[n][0]);
  return 0;
}
