You are given a positive integer n. Your task is to build a number m by flipping the minimum number of bits in the binary representation of n such that m is less than n (m < n) and it is as maximal as possible. Can you?



Input


The first line contains an integer T (1 ≤ T ≤ 105) specifying the number of test cases.

Each test case consists of a single line containing one integer n (1 ≤ n ≤ 109), as described in the statement above.



Output


For each test case, print a single line containing the minimum number of bits you need to flip in the binary representation of n to build the number m.



Example




Input

2
5
10



Output

1
2


题目意思：将一个2进制的n中每个位翻转得到一个比n小且尽可能大的数，求输出翻转了几位。 

解题思路：这道题该由我背锅，我当时先是翻译错了题意，后来稍微有一点眉目了，我又理解错了那个flip的意思，这里面的翻转并不是那种交换(swap那样的），而是像硬币正面换到反面那样的翻转，也就
是0与1的交换，根据题意可以推出想要得到一个既要比n小还有尽可能大的数，只有是n前面的那一个数n-1。所以就是根据n构造一个二进制的n-1,方法就是找到n的二进制中最后面的那一个1翻转为0，而最
后一个1之后的0全都翻转成1，统计所用的翻转次数即可。



1 #include<cstdio>
 2 #include<cstring>
 3 int main()
 4 {
 5     int t,n,j,k,i,count;
 6     int a[32];
 7     scanf("%d",&t);
 8     while(t--)
 9     {
10         scanf("%d",&n);
11         memset(a,-1,sizeof(a));
12         j=0;
13         count=0;
14         i=n;
15         while(i)
16         {
17             a[j]=i%2;
18             if(a[j]==0)
19             {
20                 count++;
21             }
22              if(a[j]==1)
23              {
24                  count++;
25                  break;
26              }
27             i/=2;
28             j++;
29         }
30         printf("%d\n",count);
31     }
32     return 0;
33 }