Given the number n, find the smallest positive integer which has exactly n divisors. It is guaranteed that for the given n the answer will not exceed 1018.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000).

Output

Output the smallest positive integer with exactly n divisors.

Examples

Input

4


Output

6


Input

6


Output

12
#!/usr/bin/env python

n = int(raw_input())

primes = [ 2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41, 
          43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101,
          103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
          173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
          241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313,
          317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
          401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467,
          479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569,
          571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643,
          647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
          739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823,
          827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911,
          919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

          
def sol(n, id, previous):
    if n == 1:
        return 1
    result = 10**18+1;
    for i in range(2, min(n, previous)+1):
        if n % i == 0:
            result = min(result, primes[id]**(i-1) * sol (n/i, id+1, i))
    return result

print (sol(n, 0, 59))