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:
import sys
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

def debug(var):
    print(var, file=sys.stderr)

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from math import gcd
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

n, = I()
a = []
b = []
for i in range(n):
    x, y = I()
    if x < y:
        a.append((x, y, i+1))
    else:
        b.append((x, y, i+1))
a.sort(reverse = 1)
b.sort()
n, m = len(a), len(b)
if n >= m:
    print(n)
    for i in a:
        print(i[2], end = ' ')
else:
    print(m)
    for i in b:
        print(i[2], end = ' ')
debug(a)
debug(b)