Dreamoon Likes Permutations
The sequence of $$$m$$$ integers is called the permutation if it contains all integers from $$$1$$$ to $$$m$$$ exactly once. The number $$$m$$$ is called the length of the permutation.Dreamoon has two permutations $$$p_1$$$ and $$$p_2$$$ of non-zero lengths $$$l_1$$$ and $$$l_2$$$.Now Dreamoon concatenates these two permutations into another sequence $$$a$$$ of length $$$l_1 + l_2$$$. First $$$l_1$$$ elements of $$$a$$$ is the permutation $$$p_1$$$ and next $$$l_2$$$ elements of $$$a$$$ is the permutation $$$p_2$$$. You are given the sequence $$$a$$$, and you need to find two permutations $$$p_1$$$ and $$$p_2$$$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
In the first example, two possible ways to divide $$$a$$$ into permutations are $$$\{1\} + \{4, 3, 2, 1\}$$$ and $$$\{1,4,3,2\} + \{1\}$$$.In the second example, the only way to divide $$$a$$$ into permutations is $$$\{2,4,1,3\} + \{2,1\}$$$.In the third example, there are no possible ways.
The first line contains an integer $$$t$$$ ($$$1 \le t \le 10\,000$$$) denoting the number of test cases in the input.Each test case contains two lines. The first line contains one integer $$$n$$$ ($$$2 \leq n \leq 200\,000$$$): the length of $$$a$$$. The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq n-1$$$).The total sum of $$$n$$$ is less than $$$200\,000$$$.
For each test case, the first line of output should contain one integer $$$k$$$: the number of ways to divide $$$a$$$ into permutations $$$p_1$$$ and $$$p_2$$$.Each of the next $$$k$$$ lines should contain two integers $$$l_1$$$ and $$$l_2$$$ ($$$1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$$$), denoting, that it is possible to divide $$$a$$$ into two permutations of length $$$l_1$$$ and $$$l_2$$$ ($$$p_1$$$ is the first $$$l_1$$$ elements of $$$a$$$, and $$$p_2$$$ is the last $$$l_2$$$ elements of $$$a$$$). You can print solutions in any order.
def count(a):
    a.sort()
    n = len(a)
    for i in range(n):
        if a[i] != i + 1:
            return 0
    return 1
t = int(input())
while t:
    n = int(input())
    a = list(map(int, input().split()))
    m = max(a)
    c1 = count(a[:m]) and count(a[m:])
    a.reverse()
    c2 = count(a[:m]) and count(a[m:])
    k = c1 + c2
    if k == 0:
        print(0)
    elif k == 1:
        print(1)
        if c1 == 1:
            print(m, n - m)
        if c2 == 1:
            print(n - m, m)
    elif m == n - m:
        print(1)
        print(m, n - m)
    else:
        print(2)
        print(m, n - m)
        print(n - m, m)
    t -= 1
import sys
read = sys.stdin.buffer.readline
for _ in range(int(read())):
    n = int(read())
    a = list(map(int, read().split()))
    ans = set()
    mx = max(a)
    for i in range(2):
        if set(a[:mx]) == set(range(1, mx + 1)) and set(a[mx:]) == set(range(1, n - mx + 1)):
            ans.add(mx)
        mx = n - mx
    print(len(ans))
    for i in ans:
        print(i, n - i)
def search(p1, p2, lp1, lp2):
    used_1 = [True] + [False] * lp1
    for ele in p1:
        if ele > lp1:
            return False
        elif used_1[ele] == False:
            used_1[ele] = True
        else:
            return False
    used_2 = [True] + [False] * lp2
    for ele in p2:
        if ele > lp2:
            return False
        elif used_2[ele] == False:
            used_2[ele] = True
        else:
            return False
    return True

def solve(ans1, ans2):
    n = int(input())
    a = [int(x) for x in input().split()]
    ma = max(a)
    count = 0
    p1 = a[0:ma:1]
    p2 = a[ma:]
    stm1 = search(p1, p2, ma, n - ma)
    if stm1:
        count += 1
        ans2.append([ma, n - ma])
    if ma == n - ma:
        ans1.append(count)
        return
    p1 = a[0:n - ma:1]
    p2 = a[n - ma:]
    stm2 = search(p1, p2, n - ma, ma)
    if stm2:
        count += 1
        ans2.append([n - ma, ma])
    ans1.append(count)
    return
n = int(input())
ans1 = []
ans2 = []
for i in range(n):
    solve(ans1, ans2)
for ele in ans1:
    print(ele)
    for i in range(ele):
        print(*ans2[0])
        ans2.pop(0)