Code Chef:cool Guys

All submissions to this problem are available.

Given an integer N. Integers A and B are chosen randomly in the range [1..N]. Calculate the probability that greatest Common Divisor (GCD) of A and b equals to b.

Input

The ' The ' input contains an integer T denoting the number of test cases. The description of T test cases follows. Each test case consists of the a single integer N on a separate line.

Output

For each test case, the output a single line containing probability as a irreducible fraction.

Given an integer N. Integers A and B are chosen randomly in the range [1..N]. Calculate the probability that greatest Common Divisor (GCD) of A and b equals to b.

Input

The ' The ' input contains an integer T denoting the number of test cases. The description of T test cases follows. Each test case consists of the a single integer N on a separate line.

Output

More Wonderful content: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

For each test case, the output a single line containing probability as a irreducible fraction.

Example

Input:
3
1
2
3

Output:
1/1
3/4
5/9

Constraints

1<=T<=10³

1<=N<=10⁹

This is also a maths problem.

There are two knowledge points:

1 Algorithm for finding GCD

2 The logarithm that is equal to the two number of a numeric value-this is a mathematical formula

It is said that here is a detailed analysis of mathematics, interested in research:

Http://matwbn.icm.edu.pl/ksiazki/mon/mon42/mon4204.pdf

OJ System:

http://www.codechef.com/problems/COOLGUYS/

 Long Long mgcd (long long A, long B) {long long c = 0;  
        while (b) {c = b;  
        b = a% B;  
    A = C;  
return C;  
    Long Long Pairsofcoolguys (long long N) {long long ans = 0;  
    Long Long sq = (long long) sqrt (n);  
    for (Unsigned i = 1; I <= sq; i++) {ans = n/i;  
    Ans = (ans<<1)-sq*sq;  
return ans;  
    } void Coolguys () {long long n = 0;  
    int T = 0;  
    cin>>t;  
        while (t--) {cin>>n;  
        Long PS = Pairsofcoolguys (n);  
        n *= N;  
        Long Long D = MGCD (PS, N);  
    cout << (ps/d) << "/" << (n/d) << "\ n"; }  
}