human
Solve in Python:
There is a building with 2N floors, numbered 1, 2, \ldots, 2N from bottom to top.
The elevator in this building moved from Floor 1 to Floor 2N just once.
On the way, N persons got on and off the elevator. Each person i (1 \leq i \leq N) got on at Floor A_i and off at Floor B_i. Here, 1 \leq A_i < B_i \leq 2N, and just one person got on or off at each floor.
Additionally, because of their difficult personalities, the following condition was satisfied:
 - Let C_i (= B_i - A_i - 1) be the number of times, while Person i were on the elevator, other persons got on or off. Then, the following holds:
 - If there was a moment when both Person i and Person j were on the elevator, C_i = C_j.
We recorded the sequences A and B, but unfortunately, we have lost some of the records. If the record of A_i or B_i is lost, it will be given to you as -1.
Additionally, the remaining records may be incorrect.
Determine whether there is a pair of A and B that is consistent with the remaining records.

-----Constraints-----
 - 1 \leq N \leq 100
 - A_i = -1 or 1 \leq A_i \leq 2N.
 - B_i = -1 or 1 \leq B_i \leq 2N.
 - All values in input are integers.

-----Input-----
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
:
A_N B_N

-----Output-----
If there is a pair of A and B that is consistent with the remaining records, print Yes.
otherwise, print No.

-----Sample Input-----
3
1 -1
-1 4
-1 6

-----Sample Output-----
Yes

For example, if B_1 = 3, A_2 = 2, and A_3 = 5, all the requirements are met.
In this case, there is a moment when both Person 1 and Person 2 were on the elevator, which is fine since C_1 = C_2 = 1.

What is a Python code to determine whether there is a pair of A and B that is consistent with the remaining records?
assistant
#Here is Python code to determine whether there is a pair of A and B that is consistent with the remaining records.

# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from fractions import Fraction
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations, accumulate
from operator import add, mul, sub, itemgetter, attrgetter

import sys
# sys.setrecursionlimit(10**6)
# readline = sys.stdin.buffer.readline
readline = sys.stdin.readline

INF = 1 << 60

def read_int():
    return int(readline())

def read_int_n():
    return list(map(int, readline().split()))

def read_float():
    return float(readline())

def read_float_n():
    return list(map(float, readline().split()))

def read_str():
    return readline().strip()

def read_str_n():
    return readline().strip().split()

def ep(*args):
    print(*args, file=sys.stderr)

def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.perf_counter()
        ret = f(*args, **kwargs)
        e = time.perf_counter()

        ep(e - s, 'sec')
        return ret

    return wrap

@mt
def slv(N, AB):
    used = set()
    for a, b in AB:
        if a > 0 and b > 0 and a >= b:
            return 'No'
        if a > 0 and a in used:
            return 'No'
        else:
            used.add(a)
        if b > 0 and b in used:
            return 'No'
        else:
            used.add(b)

    AB = [(a-1, b-1) for a, b in AB]

    @lru_cache(maxsize=None)
    def f(s):
        if s == 2*N:
            return True
        for r in range(s+2, 2*N+1, 2):
            l = r - s
            l2 = l // 2
            r2 = s + l2

            used = Counter()
            for a, b in AB:
                if a >= 0 and r2 <= a < r:
                    break
                if b >= 0 and  s <= b < r2:
                    break
                if s <= a < r2 and r2 <= b < r and b - a != l2:
                    break
                if s <= a < r2:
                   ...