In the first shot at the problem on paper, it becomes apparent that in a *Natural Latin Square*, the *trace* always turns out to be `n*(n+1)/2`

The elements in such a *main diagonal* may be either completely distinct from 1 to n or all same as the midpoint n/2 or (n+1)/2

By the end of this attempt, we arrive at no. of repeated rows

```
# Read rows and sum them
## If the sum doesn't add up to n(n+1)/2, call it a repeated row
### n^2 visits
# Read columns and sum them
## If the sum doesn't add up to n(n+1)/2, call it a repeated column
### n^2 visits again. total = 2*(n^2) visits
# if repeated_row and repeated_column are zeroes, directly print that
# the trace is n*(n+1)/2
# else sum the diagonals
# n visits again. Total = 2*(n^2)+n visits
def execute():
test_case_count = int(next(lines))
for case_num in range(test_case_count):
n = int(next(lines))
expected_sum = n*(n+1)/2
repeated_row_count = 0
repeated_col_count = 0
row_arr = []
for row_id in range(n):
row = next(lines)
row_sum = sum(map(int, row.split()))
if row_sum != expected_sum:
repeated_row_count += 1
row_arr.append(row)
print("repeated row count: " + str(repeated_row_count))
lines = open('1_in', 'r')
execute()
```