Codeforces 579A. Raising Bacteria

You are a lover of bacteria. You want to raise some bacteria in a box.

Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly x bacteria in the box at some moment.

What is the minimum number of bacteria you need to put into the box across those days?

Input

The only line containing one integer x (1 ≤ x ≤ 10⁹).

Output

The only line containing one integer: the answer.

Examples

Input

5

Output

2

Input

8

Output

1

Note

For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.

For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.

题解：题意很简单找2的次方 下面两种方法

解法一：（树状数组）

#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
#define lowbit(x) x&(-x)
int main()
{
    int n,t;
    while(cin>>n){
        t=0;
        while(n){
            t++;
            n-=lowbit(n);
        }
        cout<<t<<endl;
    }
    return 0;
}

解法二：（除以二迭代）

#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;
int main()
{
    int n,t;
    while(cin>>n){
        t=0;
        while(n){
            if(n&1) t++;
            n/=2;
        }
        cout<<t<<endl;
    }
    return 0;
}