Codeforces Round #410 (Div. 2)C题

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：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=100000+5,maxn=100000+5,inf=0x3f3f3f3f;

ll n,a[N],ans=0;

ll gcd(ll a,ll b)
{
    return b? gcd(b,a%b):a;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    ll ans=0,x;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        if(i==0)x=a[i];
        else x=gcd(x,a[i]);
        a[i]%=2;
    }
    if(x>1)
    {
        cout<<"YES"<<endl<<0<<endl;
        return 0;
    }
    for(int i=0;i<n;i++)
    {
        if(a[i]==1)
        {
            if(i+1<n)
            {
                if(a[i+1]==1)ans++,a[i+1]=0;
                else if(a[i+1]==0)ans+=2;
                a[i]=0;
            }
            else ans+=2;
        }
    }
    cout<<"YES"<<endl<<ans<<endl;
    return 0;
}