ADATREE  Ada and Trees
Ada the Ladybug is a farmer. She has a long furrow in which she grows trees. Each tree has some weigth. The task is simple, she wants to know the biggest tree on some part of the furrow which is not greater than some height H. As Ada asks for this very often, she asked you to write a program for this.
Input
The first line will contain two integer 1 ≤ N, Q ≤ 3*10^{5}, the number trees and the number of questions.
The next line will contain N integers 0 ≤ A_{i} ≤ 10^{6}, the heights of trees.
The sum next Q will contain three integers: 0 ≤ l ≤ r < N, the segment of furrow she is interested in and 0 ≤ H ≤ 10^{6}
Output
For each query output either the size of highest tree lesser/equal to H or output 0 if such tree doesn't grow on given segment.
Example Input
9 8 1 5 9 11 9 7 6 2 1 1 6 4 1 6 10 0 8 97 0 8 4 1 4 5 2 6 8 2 8 5 3 3 12
Example Output
0 9 11 2 5 7 2 11
hide comments
scolar_fuad:
20191122 05:35:23
great problem indeed


n_o_o_b:
20190911 13:44:59
Great problem!! 

DK...:
20190329 15:32:51
Solutions like O(nsqrt(n)) or O(nlog^2(n)) don't pass for me, I got AC with Persistent segment tree, it could be overkill. Last edit: 20190329 15:36:38 

DOT:
20180812 10:49:22
Square root decomposition doesn't produce the best time for me, but it does work. Maybe, try tweaking the block sizes and compare the results. Last edit: 20180812 10:50:09 

morass:
20180628 14:45:46
@rajcoolaryan: Good day to you! Can't say whether it is enough.. I would have guessed it might be, yet all judge solutions are faster than this :'( 

rajcoolaryan:
20180621 06:30:28
square root decmposition tle 

morass:
20171018 23:28:31
@shubham_001: Good day to you. No judge solution uses treap. I could imagine a solution using it, but I don't think it is the "best" way :)


shubham_001:
20171018 16:10:36
is this a treap question? 
Added by:  Morass 
Date:  20171008 
Time limit:  3s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All 
Resource:  Tunisian Collegiate Training Contest  Round #01 