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;
struct Segment {
  int x, y;
  int idx;
};
vector<Segment> ls, rs;
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    Segment s;
    scanf("%d %d", &s.x, &s.y);
    s.idx = i;
    if (s.x < s.y) {
      ls.push_back(s);
    } else {
      rs.push_back(s);
    }
  }
  vector<Segment> lpos, rpos;
  sort(ls.begin(), ls.end(),
       [](const Segment& a, const Segment& b) { return a.y > b.y; });
  Segment prev;
  for (auto s : ls) {
    if (lpos.empty() || s.x < prev.y) {
      lpos.push_back(s);
      prev = s;
    }
  }
  sort(rs.begin(), rs.end(),
       [](const Segment& a, const Segment& b) { return a.y < b.y; });
  for (auto s : rs) {
    if (rpos.empty() || s.x > prev.y) {
      rpos.push_back(s);
      prev = s;
    }
  }
  if (lpos.size() > rpos.size()) {
    printf("%d\n", (int)lpos.size());
    for (auto s : lpos) {
      printf("%d ", s.idx);
    }
  } else {
    printf("%d\n", (int)rpos.size());
    for (auto s : rpos) {
      printf("%d ", s.idx);
    }
  }
  printf("\n");
  return 0;
}