1062_B. Math
JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is as follows. Given an integer n, you can perform the following operations zero or more times:

  * mul x: multiplies n by x (where x is an arbitrary positive integer). 
  * sqrt: replaces n with √{n} (to apply this operation, √{n} must be an integer). 



You can perform these operations as many times as you like. What is the minimum value of n, that can be achieved and what is the minimum number of operations, to achieve that minimum value?

Apparently, no one in the class knows the answer to this problem, maybe you can help them?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10^6) — the initial number.

Output

Print two integers: the minimum integer n that can be achieved using the described operations and the minimum number of operations required.

Examples

Input

20


Output

10 2

Input

5184


Output

6 4

Note

In the first example, you can apply the operation mul 5 to get 100 and then sqrt to get 10.

In the second example, you can first apply sqrt to get 72, then mul 18 to get 1296 and finally two more sqrt and you get 6.

Note, that even if the initial value of n is less or equal 10^6, it can still become greater than 10^6 after applying one or more operations.
```python
import math


def sieve(n):
    mark = [0]*(n+1)
    prime = []
    for i in range(3, n+1, 2):
        if not mark[i]:
            for j in range(3*i, n+1, i+i):
                mark[j] = 1
    prime.append(2)
    for i in range(3, n+1, 2):
        if not mark[i]:
            prime.append(i)

    return prime


def ispowerof2(x):
    if x and not (x & (x-1)):
        return 0
    else:
        return 1


n = int(input())
#h = n
prime = sieve(n)
l = []

countp = 0

ans = 1
for i in prime:
    countp = 0
    if n % i == 0:
        ans *= i
        while n % i == 0:
            n //= i
            countp += 1
        l.append(countp)

    if n == 1:
        break
if len(l) != 0:
    maxp = max(l)
else:
    maxp = 0
flag = 0
for i in l:
    if i != maxp:
        flag = 1

temp = 0

if flag:
    temp = 1
else:
    temp = ispowerof2(maxp)

if maxp == 1:
    maxp -= 1
elif maxp != 0:
    maxp = math.ceil(math.log2(maxp)) + temp

print(ans, maxp)

```