Dynamic Rankings

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.

Your task is to write a program for this computer, which

- Reads N numbers from the input (1 <= N <= 50,000)

- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.

Input

The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.

The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format

Q i j k or
C i t

It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.

There're NO breakline between two continuous test cases.

Output

For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])

There're NO breakline between two continuous test cases.

Sample Input

2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3

Sample Output

3
6
3
6

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题目意思是要对一个序列询问某段当中的第k大数，并且支持修改单个元素的操作。
50000的数据范围显然要我们用高级数据结构来维护，但是很囧的问题就是：询问区间要用线段树，询问第k大要用平衡树。。。
解决这个问题的方法很简单，就是线段树套平衡树，线段树中的每个节点都有一棵平衡树，维护线段树所记录的这个区间的元素。
这样处理空间上是O(nlogn)的，因为线段树有logn层，每层的平衡树所记的节点总数都有n个。
修改很容易想到，把所有包含要修改点的区间的平衡树都修改了就行了，但查询的时候又被囧到了：询问的区间不一定就是线段树里面某个节点记的某个区间。
。。最终还是去找了hyf神牛。。。他一语点破天机：二分答案。。
对于每次查询，二分第k大的数是多少，在线段树里查询这个区间中小于等于这个值的有多少个，就可以得到答案了。
需要注意的细节是：对于一个数，可能会有重复，比如对于1 2 2 2 3，查询第2大的数，当而分到2的时候，可能查到的是第三个2，查询结果也就是4。为了解决这个问题，可以在当查询不大于x的数的个数t1时，再查询不大于x-1的数的个数t2，如果t2<k<=t1那么此时的x便是所求的第k大的数。
这样的话，查询是O(log(n)log(n)*log(MAXNUM))   (其实在查询线段树和平衡树的时候不可能同时都达到log(n)，只是具体是多少我就算不来了。。望高手解答。。)，修改是O(log(n)*log(n))的(问题同上。。)，就可以在时间范围内解决了。