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).

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 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.

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.

题解：我们发现一个位置经过两次操作a[i]变成-2a[i+1],a[i+1]变成2a[i],所以当gcd为1时我们可以把他们都变为偶数，所以我们把所有的数都变为偶数

#include<cstdio>

#include<iostream>

#include<cmath>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn=1e5+;

int a[maxn],n;

int main()

{

    scanf("%d",&n);

    for(int i=;i<n;i++)

    {

        scanf("%d",&a[i]);

    }

    int tmp=a[];

    for(int i=;i<n;i++)

    {

        tmp=__gcd(tmp,a[i]);

    }

    if(tmp!=)

    {

        puts("YES\n0");

    }

    else

    {

        int ans=;

        for(int i=;i<n;i++)

        {

            if(a[i]%==)continue;

            else if(i==n-)

            {

                ans+=;

            }

            else

            {

                if(a[i+]%!=)ans++;

                else ans+=;

                i++;

            }

        }

        printf("YES\n%d\n",ans);

    } 

}