AnInterestingSequence

YiYi is a smartboy, he is fascinated in solving eccentric problems. One day when he wasreading a book, he came across a very interesting problem. After a few time’sthinking, he finally get the solution. YiYi considers this solution veryheuristic, now he decides to show this problem to you so that you can alsoenjoy the happiness of solving this interesting problem. Here is thedescription of the problem:

Suppose at firstyou have a sequence, S1, consists of two numbers: 1, 1.According to some rule you can get the sequence Sn from thesequence Sn-1.Once you get Sn-1, you will be asked to insert one nbetween a and b if a and b are two adjacent numbers in Sn-1 and a + b = n. After you have inserted all possible n you finally get Sn.You can also get the sequence Sn+1 using the same method and so on. 

For example,S1 =1,1.Since 1 + 1 = 2, we will insert 2 between two 1s and then we get S2, S2 = 1,2,1.Now we will insert two 3s in the sequence to get S3, as you cansee, S3 = 1,3,2,3,1. In the same way, we know S4 =1,4,3,2,3,4,1, S5 = 1,5,4,3,5,2,5,3,4,5,1 and so on.

Here thequestion comes: give you n, you will be asked to output the number of n in Sn.For example if n = 5, you will output 4 since there are 4 5s in S5.

输入

The first line of the input contains a number k, the number of test cases to solve (1 ≤ k ≤ 200). Each test case consists of a single integer 2 ≤ n ≤ 10^10 on a separate line.

输出

For each test case, you are asked to output the number of n in Sn on a line.

样例输入

3245

样例输出

124

#include<stdio.h>
long long eular(long long n){
     long long ret=1,i;
  for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
        n/=i,ret*=i-1;
        while(n%i==0)
           n/=i,ret*=i;
        }
    }
   if(n>1)  ret*=n-1;
return ret;
}


int main(){
    int t;


scanf("%d",&t);


while(t--){
long long n;
scanf("%lld",&n);
printf("%lld\n",eular(n));
}
return 0;
}