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 void R(int &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = (v + (v << 2) << 1) + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
inline void R(long double &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = v * 10 + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
struct node {
  long double x, y;
} p[10005];
struct fangcheng {
  long double a, b, r;
} t[1005], sma, gian;
struct L {
  node a, b;
  L(node a, node b) : a(a), b(b) {}
  L() {}
} xian;
inline long double getdis(node a, node b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
inline long double operator*(node a, node b) { return a.x * b.y - a.y * b.x; }
inline node operator-(node a, node b) { return (node){a.x - b.x, a.y - b.y}; }
int a[100005];
inline int gcd(int x, int y) {
  if (abs(x) < abs(y)) swap(x, y);
  int t;
  while (y) {
    t = y;
    y = x % y;
    x = t;
  }
  return abs(x);
}
int main() {
  int n;
  R(n);
  int GCD;
  for (int i = 1; i <= n; ++i) {
    R(a[i]);
  }
  GCD = gcd(a[1], a[2]);
  for (int i = 3; i <= n; ++i) {
    GCD = gcd(GCD, a[i]);
  }
  if (GCD > 1) {
    cout << "YES\n0";
    return 0;
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] & 1) {
      if (a[i + 1] & 1) {
        a[i] = a[i + 1] = 2;
        ans += 1;
      } else {
        a[i] = a[i + 1] = 2;
        ans += 2;
      }
    }
  }
  cout << "YES\n";
  cout << ans;
  return 0;
}
