Problem: You are given n pairs of integers (a_1, b_1), (a_2, b_2), …, (a_n, b_n). All of the integers in the pairs are distinct and are in the range from 1 to 2 ⋅ n inclusive.

Let's call a sequence of integers x_1, x_2, …, x_{2k} good if either 

  * x_1 < x_2 > x_3 < … < x_{2k-2} > x_{2k-1} < x_{2k}, or 
  * x_1 > x_2 < x_3 > … > x_{2k-2} < x_{2k-1} > x_{2k}. 



You need to choose a subset of distinct indices i_1, i_2, …, i_t and their order in a way that if you write down all numbers from the pairs in a single sequence (the sequence would be a_{i_1}, b_{i_1}, a_{i_2}, b_{i_2}, …, a_{i_t}, b_{i_t}), this sequence is good.

What is the largest subset of indices you can choose? You also need to construct the corresponding index sequence i_1, i_2, …, i_t.

Input

The first line contains single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of pairs.

Each of the next n lines contain two numbers — a_i and b_i (1 ≤ a_i, b_i ≤ 2 ⋅ n) — the elements of the pairs.

It is guaranteed that all integers in the pairs are distinct, that is, every integer from 1 to 2 ⋅ n is mentioned exactly once.

Output

In the first line print a single integer t — the number of pairs in the answer.

Then print t distinct integers i_1, i_2, …, i_t — the indexes of pairs in the corresponding order.

Examples

Input


5
1 7
6 4
2 10
9 8
3 5


Output


3
1 5 3


Input


3
5 4
3 2
6 1


Output


3
3 2 1

Note

The final sequence in the first example is 1 < 7 > 3 < 5 > 2 < 10.

The final sequence in the second example is 6 > 1 < 3 > 2 < 5 > 4.

Solution (in PYTHON3):

Solution:
from operator import gt, lt
from collections import deque, namedtuple

n = int(input())
pairs = [[int(t) for t in input().split(' ')] for _ in range(n)]
Pair = namedtuple('Pair', ['a', 'b', 'i'])
pairs = list(Pair(*p, i) for i, p in enumerate(pairs))

def attempt(pairs, a_high):
    pairs = [p for p in pairs if a_high ^ (p.a < p.b)]
    pairs = deque(sorted(pairs, reverse=a_high, key=lambda p: p.a))
    if not pairs:
        return []

    result = [pairs.pop()]

    while pairs and (a_high ^ (result[-1].b > pairs[-1].a)):
        result.append(pairs.pop())
    
    return result


a_high = attempt(pairs, True)
a_low = attempt(pairs, False)

best = max(a_high, a_low, key=len)
print(len(best))
print(' '.join([str(p.i + 1) for p in best]))