CF-798C

C. Mike and gcd problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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. .

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.

 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 .

In the second example the gcd of the sequence is already greater than 1.

题意：

对于给定字符串，我们可将其相邻的两个字符做以下操作：

num[i],num[i+1]  ->  num[i]-num[i+1],num[i]+num[i+1]

由此可得，变换两次得：-2num[i+1],2num[i]

因为所有数均可转换为偶数，所以结果不可能为“NO”。

当相邻两数均为奇数时，只进行一次变换就可将它们全部变换为偶数；

当相邻数一奇一偶时，只要进行两次就可转换为偶数。

AC代码：

 #include<bits/stdc++.h>
 using namespace std;

 long long num[];
 int n;

 int gcd(long long a,long long b){
     if(b==){
         return abs(a);
     }
     return gcd(b,a%b);
 }

 int main(){
     cin>>n;
     for(int i=;i<n;i++){
         cin>>num[i];
     }
     long long ans=;
     for(int i=;i<n;i++){
         ans=gcd(ans,num[i]);
     }
     if(ans>){
         cout<<"YES"<<endl<<<<endl;
         return ;
     }
     ans=;
     for(int i=;i<n-;i++){
         if(num[i]&&&num[i+]&){
             ans++;
             num[i]=;
             num[i+]=;
         }
     }
     for(int i=;i<n;i++){
         if(num[i]&){
             ans+=;
         }
     }
     cout<<"YES"<<endl<<ans<<endl;

     return ;
 }