1330_B. 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.)

Input

The first line contains an integer t (1 ≤ t ≤ 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 ≤ n ≤ 200 000): the length of a. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 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 ≤ l_1, l_2 ≤ 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
import sys
input = sys.stdin.readline

Q = int(input())
Query = []
for _ in range(Q):
    N = int(input())
    A = list(map(int, input().split()))
    Query.append((N, A))

for N, A in Query:
    leftOK = [True]*(N+1)
    already = set()
    MAX = 0
    for i, a in enumerate(A):
        if a in already:
            ok = False
        else:
            already.add(a)
            MAX = max(MAX, a)
            if len(already) == i+1 and MAX == i+1:
                ok = True
            else:
                ok = False
        leftOK[i+1] = ok
    
    rightOK = [True]*(N+1)
    already = set()
    MAX = 0
    for i in reversed(range(N)):
        a = A[i]
        if a in already:
            ok = False
        else:
            already.add(a)
            MAX = max(MAX, a)
            if len(already) == MAX and MAX == N-(i):
                ok = True
            else:
                ok = False
        rightOK[i] = ok

    ans = []
    for n in range(1, N):
        if leftOK[n] and rightOK[n]:
            ans.append(str(n) + " " + str(N-n))
    
    print(len(ans))
    print("\n".join(ans))
```