Let's assume that

• v(n) is the largest prime number, that does not exceed n;
• u(n) is the smallest prime number strictly greater than n.

Find .

The first line contains integer t (1 ≤ t ≤ 500) — the number of testscases.

Each of the following t lines of the input contains integer n (2 ≤ n ≤ 109).

Print t lines: the i-th of them must contain the answer to the i-th test as an irreducible fraction "p/q", where p, q are integers, q > 0.

2
2
3

1/6
7/30

分析：把公式分解1/(p*q)=1/(p-q) *  (1/q-1/p) 然后求和发现公式：ans=(-2q+2*n-2*p+2+2*q*q)/(2*u*v);

 1 #include<cstdio>

 2 #include<cmath>

 3 #include<algorithm>

 4 using namespace std;

 5 bool isprime(unsigned long long x)

 6 {

 7     int idx=sqrt(x);

 8     for(int i=; i<=idx; ++i)

 9         if(x%i==)

             return false;

     return true;

 }

 int main()

 {

     int t;

     unsigned long long n;

     scanf("%d",&t);

     while(t--)

     {

         scanf("%I64u",&n);

         unsigned long long v=n,u=n+;

         while(!isprime(v))

             --v;

         while(!isprime(u))

             ++u;

         unsigned long long p=v*u-*u+*n-*v+,q=*v*u,tmp=__gcd(p,q);

         printf("%I64u/%I64u\n",p/tmp,q/tmp);

     }

 }