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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = (int)1e9 + 7;
const int N = (int)3e5 + 7;
vector<pair<pair<int, int>, int> > p[2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > y) {
      p[0].push_back(make_pair(make_pair(y, x), i));
    } else {
      p[1].push_back(make_pair(make_pair(y, x), i));
    }
  }
  sort((p[0]).begin(), (p[0]).end());
  sort((p[1]).begin(), (p[1]).end());
  reverse((p[1]).begin(), (p[1]).end());
  if ((int)(p[0]).size() > (int)(p[1]).size()) {
    printf("%d\n", (int)(p[0]).size());
    for (int i = (0); i < ((int)(p[0]).size()); ++i) {
      printf("%d ", p[0][i].second);
    }
  } else {
    printf("%d\n", (int)(p[1]).size());
    for (int i = (0); i < ((int)(p[1]).size()); ++i) {
      printf("%d ", p[1][i].second);
    }
  }
  printf("\n");
  return 0;
}