For those who want the editorial sooner:

```
from collections import defaultdict
for _ in range(int(input())):
n, q = map(int, input().split())
# defaultdict(list) assigns the value as an empty list for each key by default
# ind stores a unique value for each range
ind = defaultdict(list)
# Unique value for each range
unique_val = 1
for _ in range(q):
l, r = map(int, input().split())
# Converting to 0-based indexing
l -= 1
r -= 1
# Positive denotes beginning
ind[l].append(unique_val)
# Negative denotes ending
# We use r+1 because r is inclusive in the range
ind[r+1].append(-unique_val)
# Next value to make it unique
unique_val += 1
no_sts = 0 # Number of statues that will be destroyed
ispeed = 0 # Speed at which the number of statues destroyed increases
# First occurrence of a range (stores l value of a range)
occ = {}
for i in range(n):
# For each "endpoint" (which is l or r+1)
for j in ind[i]:
# If it's a new range the speed with which the number of statues is destroyed will increase
if j > 0:
occ[j] = i
ispeed += 1
# If it's the end of a range
# The speed with which the number of statues is destroyed will decrease
# The current amount of statues which are being destroyed at each index will decrease by how much it has contributed
# Which is equal to r - l + 1, but since we set the endpoint to r+1, we should make it r - l
# Here, (r = i) and (l = [first occurrence of -j] = occ[-j])
elif j < 0:
ispeed -= 1
no_sts -= i - occ[-j]
# The amount of statues destroyed at each index increases by the speed
no_sts += ispeed
print(no_sts, end=' ')
print()
```

Pastebin link: Angry Cyborg - Pastebin.com

CodeChef submission link: Solution: 40945043 | CodeChef

Time complexity: O(N+Q)

Space complexity: O(N+Q)

~~A C++ solution is under construction!~~ C++ version here!

If you have any queries/suggestion, please let me know!