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.

题意：给一串数，求gcd，如果gcd==1，那么可以通过改变a[i],a[i+1]，变成a[i]-a[i+1],a[i]+a[i+1]，求最小改变次数

题解： 比赛时没做出来，一直以为要用gcd模拟！（真是越来越蠢，忘了数论题推公式很重要了！）

证明：设d是改变后的gcd，d|a[i]-a[i+1],d|a[i]+a[i+1],得d|2*a[i],d|2*a[i+1],

那么d|gcd(a[0],...2*a[i],2*a[i+1],...,a[n-1]),d|2*gcd(a[0],...a[i],a[i+1],...,a[n-1])

而gcd(a[0],...a[i],a[i+1],...,a[n-1])==1，则d==2；

最后通过把每个数%2计算总和

#include<map>

#include<set>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

using namespace std;

const int N=+,maxn=+,inf=0x3f3f3f3f;

ll n,a[N],ans=;

ll gcd(ll a,ll b)

{

    return b? gcd(b,a%b):a;

}

int main()

{

    ios::sync_with_stdio(false);

    cin.tie();

    cin>>n;

    ll ans=,x;

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

    {

        cin>>a[i];

        if(i==)x=a[i];

        else x=gcd(x,a[i]);

        a[i]%=;

    }

    if(x>)

    {

        cout<<"YES"<<endl<<<<endl;

        return ;

    }

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

    {

        if(a[i]==)

        {

            if(i+<n)

            {

                if(a[i+]==)ans++,a[i+]=;

                else if(a[i+]==)ans+=;

                a[i]=;

            }

            else ans+=;

        }

    }

    cout<<"YES"<<endl<<ans<<endl;

    return ;

}