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 <image>.

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.

Examples

Input

2
2
3


Output

1/6
7/30
T = int( input() )

MAX = 33000

bePrime = [0] * MAX;
primNum = []

for j in range(2, MAX):
    if bePrime[j] == 0: 
        primNum.append( j )
        i = j
        while i < MAX:
            bePrime[i] = 1
            i = i + j

def isPrime( a ):
    for j in primNum:
        if j >= a:
            return True
        if a % j == 0:
            return False
    return True

def gcd( a, b ):
    if b == 0:
        return a
    return gcd( b, a % b );

while T > 0:
    num = 0;
    n = int( input() )

    m = n
    while isPrime(m) == False:
        m -= 1
    while isPrime(n + 1) == False:
        n += 1
        num += 1

    a = n - 1
    b = 2 * ( n+1 )

    a = a * (n+1) * m - num * b
    b = b * (n+1) * m

    g = gcd( a, b)
    a //= g
    b //= g


    print( '{0}/{1}'.format( a, b  ) )
    T -= 1;

