Navi got a task at school to collect N stones. Each day he can collect only one stone. As N can be a very large number so it could take many days to complete the task, but then he remembers that his mother gave him a magic that can double anything (i.e if he has 2 stones, the magic will make them to 4 stones). Navi can use this magic any number of time on the collected stone on a particular day and add this to the previously collected stones. Remember that he wants exactly N stones and he can't throw any stone. If he gets more than N stones then he gets 0 marks, of course he doesn't want 0 marks. Help him to collect exactly N stones in minimum number of days.
Input:
First line of input will contain number of test cases (T). Then next T lines contains a single number N, which is number of stones Navi has to collect.
Output:
For each test case, Print a single number which is the minimum number of days taken by Navi to complete the task.
Constraints:
1 <= T <= 10^5
0 <= N <= 10^9
Sample Input
2
1
3
Sample Output
1
2
Explanation
In the second case when n = 3, he will collect 1 stone on the first day and double it and collect the remaining one on the next day.
C Implementation:
Input:
First line of input will contain number of test cases (T). Then next T lines contains a single number N, which is number of stones Navi has to collect.
Output:
For each test case, Print a single number which is the minimum number of days taken by Navi to complete the task.
Constraints:
1 <= T <= 10^5
0 <= N <= 10^9
Sample Input
2
1
3
Sample Output
1
2
Explanation
In the second case when n = 3, he will collect 1 stone on the first day and double it and collect the remaining one on the next day.
C Implementation:
/* logic is counting the number of ones*/
#include <stdio.h>
int ones(int n)
{
int count = 0;
while(n)
{
if(n&1)
count++;
n>>=1;
}
return count;
}
int main()
{
int t;
int n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("%d\n",ones(n));
}
return 0;
}