UPDATEIT - Update the array !

You have an array containing n elements initially all 0. You need to do a number of update operations on it. In each update you specify l, r and val which are the starting index, ending index and value to be added. After each update, you add the 'val' to all elements from index l to r. After 'u' updates are over, there will be q queries each containing an index for which you have to print the element at that index.

Input

First line consists of t, the number of test cases. (1 ≤ t ≤ 10)

Each test case consists of "n u", number of elements in the array and the number of update operations, in the first line (1 ≤ n ≤ 10000 and 1 ≤ u ≤ 100000)

Then follow u lines each of the format "l r val" (0 ≤ l, r < n, 0 ≤ val ≤ 10000)

Next line contains q, the number of queries. (1 ≤ q ≤ 10000)

Next q lines contain an index (0 ≤ index < n)

Output

For each test case, output the answers to the corresponding queries in separate lines.

Example

Input:
1
5 3
0 1 7
2 4 6
1 3 2
3
0
3
4

Output:
7
8
6

Added by:Pandian
Date:2013-10-15
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:Own

hide comments
2018-08-09 22:54:54
Did anyone solve using bitmask?
2018-08-09 22:53:07
sqrt decomposition or BIT or difference array
2018-08-07 18:40:10
difference array seems best for this :)
2018-07-05 03:43:28
damn printf scanf
2018-07-03 06:50:32
done using priority_queues :-)
2018-06-19 17:09:46
bit does the job simple implementation .
2018-06-19 12:37:03
5 lines of code and 5 TLE due to cin cout..
2018-06-18 21:33:42
No BIT or segment tree needed
2018-05-09 08:14:53
how to read and write to standard input output (console) in Javascript (Spidermonkey , Rhino). I tried using readline() but it does not work.
2017-12-22 21:44:37
Difference array technique gives overall O(n) .
O(1) ->For each update
O(n) -> Processing
O(1) ->Query
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