cf 798c Mike and gcd problem

. 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.

#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
int gcd(int a,int b)
{
    if(b) while((a%=b) && (b%=a));
    return a+b;
}
int main()
{
    int n;
    int num[100005];
    memset(num,0,sizeof(num));
    cin >> n;
    for(int i=0;i<n;i++){
        cin  >> num[i];
    }
    int temp;
    for(int i=0;i<n;i++){
        if(i==0){
            temp = gcd(num[i],num[i+1]);
        }
        else{
            temp = gcd(temp,num[i]);
        }
    }
    if(temp > 1){
        cout << "YES" << endl;
        cout << 0 << endl;
    }
    else{
        int cnt = 0;
        for(int i=0;i<n-1;i++){
            if(num[i]&1==0 && num[i+1]&1==0) continue;
            if(num[i]&1 && num[i+1]&1){
                num[i+1] = num[i]+num[i+1];
                cnt++;
                continue;
            }
            else if(num[i]&1){
                num[i+1] = 2;
                cnt += 2;
                continue;
            }
        }
        if(num[n-1]&1) cnt+=2;
        cout << "YES" << endl;
        cout << cnt << endl;
    }
}