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>
#pragma GCC optimize("O3")
using namespace std;
int A[1 << 20], B[1 << 20], n;
vector<int> solve1(vector<int> I) {
  vector<int> ret;
  sort(I.begin(), I.end(),
       [](const int &a, const int &b) { return B[a] > B[b]; });
  int prv = INT_MAX;
  for (int i : I) {
    if (A[i] < prv) {
      ret.push_back(i);
      prv = B[i];
    }
  }
  return ret;
}
vector<int> solve2(vector<int> I) {
  vector<int> ret;
  sort(I.begin(), I.end(),
       [](const int &a, const int &b) { return B[a] < B[b]; });
  int prv = 0;
  for (int i : I) {
    if (A[i] > prv) {
      ret.push_back(i);
      prv = B[i];
    }
  }
  return ret;
}
int main(int argc, char **argv) {
  scanf("%d", &n);
  vector<int> t1, t2;
  for (int i = 0; i < (n); i++) {
    scanf("%d %d", A + i, B + i);
    if (A[i] < B[i])
      t1.push_back(i);
    else
      t2.push_back(i);
  }
  t1 = solve1(t1);
  t2 = solve2(t2);
  vector<int> ans = t1;
  if (((int)t2.size()) > ((int)t1.size())) ans = t2;
  printf("%d\n", ((int)ans.size()));
  for (int i = 0; i < (((int)ans.size())); i++)
    printf("%d%c", ans[i] + 1, " \n"[i == ((int)ans.size()) - 1]);
  return 0;
}