CodeForce-798C Mike and gcd problem（贪心）

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.

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.

题意:n个数,n<=1e5,操作:把a[i],a[i+1] 替换成 a[i]-a[i+1],a[i]+a[i+1],问至少要多少次操作才能让整个a数组的最大公约数gcd大于1.

由题目给出操作可知：当gcd(a,b)<=1时，进行操作为：

初始：a  b

第一步：a-b  a+b

第二步：-2b  2a

即两个数最多2步操作就能满足GCD==2。

对于两个偶数，要进行0步操作；对于两个奇数，要进行1步操作；对于一个奇数一个偶数，要进行2步操作。

先把所有“2个奇数成对”的情况计数+1并把两个奇数更新为偶数，然后在重新判断所有“1个奇数1个偶数成对”的情况计数+2。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[200050],n,num=0;
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++)
    scanf("%lld",&a[i]);
    ll ans=gcd(abs(a[1]),abs(a[2]));
    for(int i=3;i<=n;i++)\
        ans=gcd(ans,abs(a[i]));
    if(ans>1) cout<<"YES"<<endl<<0<<endl;
    else
    {
        for(int i=1;i<n;i++)
        if(a[i]%2&&a[i+1]%2)
            a[i]=0,a[i+1]=0,num++;
        for(int i=1;i<n;i++)
        if((a[i]%2&&a[i+1]%2==0)||(a[i]%2==0&&a[i]%2))
            a[i]=0,a[i+1]=0,num+=2;
        cout<<"YES"<<endl<<num<<endl;
    }
    return 0;
}