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 bisect import bisect_left
import sys

def solve(a, cmp):
    # if not a:
    #     return []
    # print(up)
    res = [-1 for _ in a]
    ptr = 0
    for id, e in enumerate(a):
        if ptr == 0 or cmp(a[res[ptr - 1]][0][1], e[0][0]):
            res[ptr] = id
            ptr += 1
    return [a[res[id]][1] for id in range(ptr)]


n = int(input())
a = [tuple(int(e) for e in line.split()) for line in sys.stdin]

up = [(e, id) for id, e in enumerate(a) if e[0] < e[1]]
down = [(e, id) for id, e in enumerate(a) if e[0] > e[1]]


up.sort(key=lambda e: -e[0][1])
down.sort(key=lambda e: e[0][1])

res = solve(up, lambda a, b: a > b)
res_down = solve(down, lambda a, b: a < b)

if len(res) < len(res_down):
    res = res_down

print(len(res))
print(" ".join(str(_ + 1) for _ in res))