codeforces 798C Mike and gcd problem

C.Mike and gcd problem 

Mike has a sequence A = [a₁, a₂, ..., a_n] of length n. He considers the sequence B = [b₁, b₂, ..., b_n] 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 a_i, a_i + 1 and put numbers a_i - a_i + 1, a_i + a_i + 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 b_i 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, ..., a_n (1 ≤ a_i ≤ 10⁹) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence Abeautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequenceA beautiful.

Example

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.

解题思路：

题意为：输入给定数字，求他们的大于1的最大公约数，如果没有则进行操作

对数字a,b进行操作: a,b  a-b,a+b  -2b,2a

由上可知，不论两个数字为什么，最多操作两次就有公约数2

那么有三种情况：1.都为偶数，不需要进行操作。2.一个为奇数一个为偶数，需要操作两次，3.都为奇数，只需操作一次；

然后随便打下就行了

实现代码：

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

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    ll m,a[100005],i,ans,num;
   cin>>m;
   num = 0;
   for(i=1;i<=m;i++)
    cin>>a[i];
   ans = gcd(abs(a[1]),abs(a[2]));
   for(i=3;i<=m;i++)
    ans = gcd(ans,abs(a[i]));
   if(ans>1)
    cout<<"YES"<<endl<<"0"<<endl;
   else
   {
       for(i=1;i<m;i++)

           if(a[i]%2&&a[i+1]%2)
           {
               a[i]=0;a[i+1]=0;num++;
           }
        for(i=1;i<m;i++)
           if((a[i]%2==0&&a[i+1]%2)||(a[i]%2&&a[i+1]%2==0))
            {
               a[i]=0;a[i+1]=0;num+=2;
            }

       cout<<"YES"<<endl<<num<<endl;
   }
   return 0;
}