Codeforces Round #232 (Div. 2) D. On Sum of Fractions

D. On Sum of Fractions

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 .

Input

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).

Output

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.

Sample test(s)

input

2
2
3

output

1/6
7/30

typedef long long LL ;

bool  is_prime(LL x){
      for(LL i = 2 ; i * i <= x ; i++)
        if(x % i == 0)
           return 0 ;
      return 1 ;
}

LL   V(LL x){
      while(! is_prime(x))
         x-- ;
      return x ;
}

LL   U(LL x){
      x++ ;
      while(! is_prime(x))
          x++ ;
      return x ;
}

LL   gcd(LL x , LL y){
     return y == 0 ? x : gcd(y , x % y)  ;
}

class Node{
    public :
       LL zi ;
       LL mu ;
    public :
       Node(){} ;
       Node(LL z , LL m){
           LL g = gcd(z , m) ;
           zi = z/g ;
           mu = m/g ;
       } ;
       Node  operator + (const Node &other){
              LL  m  , z , g ;
              g = gcd(mu , other.mu) ;
              m = mu / g * other.mu ;
              z = other.mu / g * zi + mu /g * other.zi ;
              g = gcd(z, m)  ;
              return Node(z/g , m/g) ;
       }
       Node  operator - (const Node &other){
              LL  m  , z  , g ;
              g = gcd(mu , other.mu) ;
              m = mu / g * other.mu ;
              z = other.mu /g * zi - mu / g * other.zi ;
              g = gcd(z, m)  ;
              return Node(z/g , m/g) ;
       }

       Node & operator = (const Node &now){
              this->mu = now.mu ;
              this->zi = now.zi ;
              return *this ;
       }
       friend ostream & operator << (ostream &out , const Node &A){
              out<<A.zi<<"/"<<A.mu ;
              return out ;
       }
};

int main(){
    int t  ;
    LL x ;
    cin>>t ;
    while(t--){
        cin>>x ;
        LL v = V(x) ;
        LL u = U(x) ;
        Node ans = Node(1 , 2) - Node(1 , v) ;
        Node  sum = Node(x-v+1, u*v) + ans ;
        cout<<sum<<endl ;
    }
    return 0;
}