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.)


-----Input-----

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 $200\,000$.


-----Output-----

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.


-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1

Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0



-----Note-----

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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'def per(X):\n    S=set(X)\n    if not len(X)==len(S):\n        return False\n    for i in range(1,len(X)+1):\n        if i not in S: return False\n    return True\nfor y in range(int(input())):\n    n=int(input())\n    L=list(map(int,input().split()))\n    m=max(L)\n    r=[]\n    if n!=m:\n        if per(L[:m]) and per(L[m:]):\n            r.append((m,n-m))\n        if per(L[-m:]) and per(L[:-m]):\n            r.append((n-m,m))\n    r=list(set(r))\n    print(len(r))\n    for a,b in r:\n        print(a,b)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13513