Codeforces Round #410 (Div. 2) Mike and gcd problem

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, ..., 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 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.

分析：首先我们可以扫一遍看看 gcd是否已经大于1不大于1的话就需要继续构造，构造gcd为2，即全为偶数

连续的两个奇数进行一次操作就变成了 两个偶数；

连续的一奇一偶进行2次操作变成两个偶数；

代码如下：

#include <bits/stdc++.h> 
using namespace std;
int c[100100];
int main()
{
    int n,h,numa,numb,cnt;
    while(cin>>n)
    {
        cnt=0;
        numa=0;
        numb=0;
        for(int i=0;i<n;i++)
        cin>>c[i];
        h=c[0];
        for(int i=0;i<n;i++)
        {
            if(c[i]%2==1)
            numa++;
            else
            numb++;
        //    cout<<" "<<c[i]<<endl;
            h=__gcd(h,c[i]);
        //    cout<<h<<endl;
        }
    //    h=__gcd(h,c[n-1]);
        if(h>1)
        {
            puts("YES");
            cout<<"0"<<endl;
            continue;
        }
        else
        {
            for(int i=0;i<n-1;i++)
            {
              if(c[i]%2==1)
              {
                    if(c[i+1]%2==1)
                    cnt+=1;
                    else
                    cnt+=2;
                    c[i]=2;
                    c[i+1]=2;
              }    
            }
            if(c[n-1]%2==1)
            cnt+=2;
            cout<<"YES"<<endl;
            cout<<cnt<<endl;
        }
    }
}