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
from fractions import gcd
def _try_composite(a, d, n, s):
    if pow(a, d, n) == 1:
        return False
    for i in range(s):
        if pow(a, 2**i * d, n) == n-1:
            return False
    return True # n  is definitely composite
 
def is_prime(n, _precision_for_huge_n=16):
    if n in _known_primes or n in (0, 1):
        return True
    if any((n % p) == 0 for p in _known_primes):
        return False
    d, s = n - 1, 0
    while not d % 2:
        d, s = d >> 1, s + 1
    # Returns exact according to http://primes.utm.edu/prove/prove2_3.html
    if n < 1373653: 
        return not any(_try_composite(a, d, n, s) for a in (2, 3))
    if n < 25326001: 
        return not any(_try_composite(a, d, n, s) for a in (2, 3, 5))
    if n < 118670087467: 
        if n == 3215031751: 
            return False
        return not any(_try_composite(a, d, n, s) for a in (2, 3, 5, 7))
    if n < 2152302898747: 
        return not any(_try_composite(a, d, n, s) for a in (2, 3, 5, 7, 11))
    if n < 3474749660383: 
        return not any(_try_composite(a, d, n, s) for a in (2, 3, 5, 7, 11, 13))
    if n < 341550071728321: 
        return not any(_try_composite(a, d, n, s) for a in (2, 3, 5, 7, 11, 13, 17))
    # otherwise
    return not any(_try_composite(a, d, n, s) 
                   for a in _known_primes[:_precision_for_huge_n])
_known_primes = [2, 3]
_known_primes += [x for x in range(5, 1000, 2) if is_prime(x)]

t = input()
while(t>0):
    t-=1
    n=input()
    p1=n;
    p2=n+1;
    while(not is_prime(p1)):
        p1-=1
    while(not is_prime(p2)):
        p2+=1
    num=(p1*p2)-(2*p2)+(2*n)-(2*p1)+2
    den=2*p1*p2
    g=gcd(num,den)
    num/=g
    den/=g
    print str(num)+"/"+str(den)


