Lucky Permutation
You are given a permutation$$$^\dagger$$$ $$$p$$$ of length $$$n$$$.In one operation, you can choose two indices $$$1 \le i < j \le n$$$ and swap $$$p_i$$$ with $$$p_j$$$.Find the minimum number of operations needed to have exactly one inversion$$$^\ddagger$$$ in the permutation.$$$^\dagger$$$ A permutation is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).$$$^\ddagger$$$ The number of inversions of a permutation $$$p$$$ is the number of pairs of indices $$$(i, j)$$$ such that $$$1 \le i < j \le n$$$ and $$$p_i > p_j$$$.
In the first test case, the permutation already satisfies the condition.In the second test case, you can perform the operation with $$$(i,j)=(1,2)$$$, after that the permutation will be $$$[2,1]$$$ which has exactly one inversion.In the third test case, it is not possible to satisfy the condition with less than $$$3$$$ operations. However, if we perform $$$3$$$ operations with $$$(i,j)$$$ being $$$(1,3)$$$,$$$(2,4)$$$, and $$$(3,4)$$$ in that order, the final permutation will be $$$[1, 2, 4, 3]$$$ which has exactly one inversion.In the fourth test case, you can perform the operation with $$$(i,j)=(2,4)$$$, after that the permutation will be $$$[2,1,3,4]$$$ which has exactly one inversion.
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).The second line of each test case contains $$$n$$$ integers $$$p_1,p_2,\ldots, p_n$$$ ($$$1 \le p_i \le n$$$). It is guaranteed that $$$p$$$ is a permutation.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case output a single integer — the minimum number of operations needed to have exactly one inversion in the permutation. It can be proven that an answer always exists.
def solve():
    n = int(input())
    a = list(map(int, input().split()))
    a = list(map(lambda x: x - 1, a))
    par = [-1 for i in range(n)]
    for i in range(n):
        if par[i] != -1:
            continue
        par[i] = i
        j = a[i]
        while j != i:
            par[j] = i
            j = a[j]
    c = set()
    for i in par:
        c.add(i)
    ans = n - len(c)
    f = 1
    for i in range(n - 1):
        if par[i] == par[i + 1]:
            f = -1
            break
    print(ans + f)
    return
if __name__ == '__main__':
    t = int(input())
    while t > 0:
        t -= 1
        solve()
def ss(a, n):
    for i in range(n):
        a[i] = a[i] - 1
    visited = [False for i in range(n)]
    t_visited = [False for i in range(n)]
    total = 0
    true = False
    for i in range(n):
        if visited[i] == False and i != a[i]:
            count = 0
            x = i
            while visited[x] == False:
                visited[x] = True
                t_visited[x] = True
                if true == False:
                    if x > 0 and t_visited[x - 1] == True:
                        true = True
                    if x < n - 1 and t_visited[x + 1] == True:
                        true = True
                x = a[x]
                count += 1
            if true == False:
                x = i
                while t_visited[x] == True:
                    t_visited[x] = False
                    x = a[x]
            total += count - 1
    if true == True:
        return total - 1
    return total + 1
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    ans = ss(a, n)
    print(ans)
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    d = [0] * n
    c = 0
    for i in range(n):
        if d[a[i] - 1] == 0:
            c += 1
            t = i
            while d[t] == 0:
                d[t] = c
                t = a[t] - 1
    ans = n - c + 1
    for i in range(n - 1):
        if d[i] == d[i + 1]:
            ans -= 2
            break
    print(ans)