You are given an array A of N positive integers. 

In one operation, you can do the following:
Choose an index i (1 ≤ i ≤ N), and change the value of A_{i} to either (A_{i})^2 or \lfloor \sqrt{A_{i}} \rfloor.

Find the minimum number of operations required, such that, for the final array A:
A is sorted in non-decreasing order;
A_{N} ≤ 10^{18}.

NOTE: The elements of the array might not fit in a 32-bit integer.

------ Input Format ------ 

- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of two lines of input:
- The first line of each test case contains N - the size of the array.
- The next line contains N integers, A_{1}, A_{2}, A_{3}, \ldots, A_{N} - the elements of the array.

------ Output Format ------ 

For each test case, output on a new line, the minimum number of operations required.

------ Constraints ------ 

$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 10^{5}$
$1 ≤ A_{i} ≤ 10^{18}$
- The sum of $N$ over all test cases won't exceed $10^{5}$.

----- Sample Input 1 ------ 
3
4
1 3 9 9
4
1 9 3 9
3
4 2 1

----- Sample Output 1 ------ 
0
1
3

----- explanation 1 ------ 
Test case $1$: The array is already sorted, no operation is required.

Test case $2$: Using one operation, we can change $A_{2}$ to $\lfloor \sqrt{A_{2}}\rfloor = 3$. The final array is $A = [1, 3, 3, 9]$ which is sorted in non-decreasing order.

Test case $3$: We make the following operations:
- Change $A_{2}$ to $\lfloor \sqrt{A_{2}}\rfloor = 1$.
- Change $A_{1}$ to $\lfloor \sqrt{A_{1}}\rfloor = 2$.
- Change $A_{1}$ to $\lfloor \sqrt{A_{1}}\rfloor = 1$.

The final array is $A = [1, 1, 1]$ which is sorted in non-decreasing order.

Difficulty: VERY_HARD
Tags: Graph algorithms

Graph traversal

Sorting

Amortized analysis

Complete search

Dynamic programming
Skill types: Dynamic programming

Sorting

Amortized analysis

Complete search

Starter code:

Solutions:
from collections import deque
import math

def getsqrt(x):
	s = int(math.sqrt(x))
	while (s + 1) * (s + 1) <= x:
		s += 1
	while s * s > x:
		s -= 1
	return s

def bfs(x):
	queue = [[x, 0]]
	reached = set([x])
	for (u, d) in queue:
		if u <= 10 ** 9 and u * u not in reached:
			reached.add(u * u)
			queue.append([u * u, d + 1])
		y = getsqrt(u)
		if y not in reached:
			reached.add(y)
			queue.append([y, d + 1])
	return sorted(queue)
for _ in range(int(input())):
	n = int(input())
	a = list(map(int, input().split()))
	prv = [[1, 0]]
	for x in a:
		costs = bfs(x)
		dp = []
		(ptr, mn) = (0, 10 ** 18)
		for (y, d) in costs:
			while ptr < len(prv) and prv[ptr][0] <= y:
				mn = min(mn, prv[ptr][1])
				ptr += 1
			dp.append([y, d + mn])
		prv = dp
	ans = 10 ** 18
	for (x, d) in prv:
		ans = min(ans, d)
	print(ans)


def mi():
	return map(int, input().split())

def li():
	return list(mi())

def si():
	return str(input())

def ni():
	return int(input())
from math import sqrt

def getsqrt(x):
	s = int(sqrt(x))
	while (s + 1) * (s + 1) <= x:
		s += 1
	while s * s > x:
		s -= 1
	return s

def genall(x):
	queue = [[x, 0]]
	reached = set([x])
	for (u, d) in queue:
		if u <= 10 ** 9 and u * u not in reached:
			reached.add(u * u)
			queue.append([u * u, d + 1])
		y = getsqrt(u)
		if y not in reached:
			reached.add(y)
			queue.append([y, d + 1])
	return sorted(queue)
for _ in range(int(input())):
	n = int(input())
	a = list(map(int, input().split()))
	prv = [[1, 0]]
	for x in a:
		costs = genall(x)
		dp = []
		(ptr, mn) = (0, 10 ** 18)
		for (y, d) in costs:
			while ptr < len(prv) and prv[ptr][0] <= y:
				mn = min(mn, prv[ptr][1])
				ptr += 1
			dp.append([y, d + mn])
		prv = dp
	ans = 10 ** 18
	for (x, d) in prv:
		ans = min(ans, d)
	print(ans)