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;
const int MAX_N = 100002;
int n, out, pos1, pos2, c;
int a[MAX_N], b[MAX_N];
stack<int> pos;
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    b[i] = a[i] % 2;
  }
  c = a[0];
  for (int i = 1; i < n; i++) {
    c = gcd(c, a[i]);
  }
  for (int i = 0; i < n; i++) {
    if (b[i] == 1) {
      pos.push(i);
    }
  }
  while (!pos.empty()) {
    if (pos.size() == 1) {
      out += 2;
      pos.pop();
    } else {
      pos2 = pos.top();
      pos.pop();
      pos1 = pos.top();
      pos.pop();
      if (pos1 + 1 == pos2) {
        out += 1;
      } else {
        out += 2;
        pos.push(pos1);
      }
    }
  }
  if (c == 1) {
    printf("YES\n%d\n", out);
  } else {
    printf("YES\n%d\n", 0);
  }
}
