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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("avx,avx2")
using namespace std;
const int mod = 1e9 + 7, q = 1e5 + 7;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
bool ok(vector<long long>& a) {
  long long g = 0;
  for (auto x : a) {
    g = gcd(g, abs(x));
  }
  return g > 1;
}
void print(vector<long long>& a) {
  for (long long x : a) cout << x << " ";
  cout << "\n";
}
void p1(vector<long long>& a, long long& cnt) {
  for (int i = 0; i < (int)(a.size()) - 1; ++i) {
    if (abs(a[i] % 2) == 1 && abs(a[i + 1] % 2) == 1) {
      long long b = a[i];
      a[i] = a[i] - a[i + 1];
      a[i + 1] += b;
      ++cnt, ++i;
      continue;
    }
  }
}
void p2(vector<long long>& a, long long& cnt) {
  for (int i = 0; i < (int)(a.size()) - 1; ++i) {
    if (abs(a[i] % 2) != abs(a[i + 1] % 2)) {
      long long b = a[i];
      a[i] = a[i] - a[i + 1];
      a[i + 1] += b;
      ++cnt, ++i;
      continue;
    }
  }
}
int main() {
  cin.tie(0), cout.tie(0), ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<long long> a(n);
  long long g = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    g = gcd(a[i], g);
  }
  if (g > 1) {
    cout << "Yes\n0";
    exit(0);
  }
  long long cnt = 0, bad = 0;
  for (int i = 0; i < 10; ++i) {
    p1(a, cnt);
    p2(a, cnt);
  }
  if (ok(a)) {
    cout << "YES\n" << cnt;
  } else
    cout << "NO";
  return 0;
}
