CF-798C

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：

对于给定字符串，我们可将其相邻的两个字符做以下操作：

num[i],num[i+1]  ->  num[i]-num[i+1],num[i]+num[i+1]

由此可得，变换两次得：-2num[i+1],2num[i]

因为所有数均可转换为偶数，所以结果不可能为“NO”。

当相邻两数均为奇数时，只进行一次变换就可将它们全部变换为偶数；

当相邻数一奇一偶时，只要进行两次就可转换为偶数。

AC代码：

#include<bits/stdc++.h>
using namespace std;

long long num[100010];
int n;

int gcd(long long a,long long b){
    if(b==0){
        return abs(a);
    }
    return gcd(b,a%b);
}

int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>num[i];
    }
    long long ans=0;
    for(int i=0;i<n;i++){
        ans=gcd(ans,num[i]);
    }
    if(ans>1){
        cout<<"YES"<<endl<<0<<endl;
        return 0;
    }
    ans=0;
    for(int i=0;i<n-1;i++){
        if(num[i]&1&&num[i+1]&1){
            ans++;
            num[i]=2;
            num[i+1]=2;
        }
    }
    for(int i=0;i<n;i++){
        if(num[i]&1){
            ans+=2;
        }
    }
    cout<<"YES"<<endl<<ans<<endl;
    
    return 0;
}