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;
bool comp(pair<int, int> x, pair<int, int> y) { return x.first > y.first; }
bool compd1(pair<int, int> x, pair<int, int> y) { return x.first < y.first; }
int main() {
  int n;
  cin >> n;
  int a[n], b[n], ta = 0, td = 0, t, ba[n], bd[n];
  vector<pair<int, int> > aa, ad;
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    if (a[i] < b[i]) {
      ta++;
      aa.push_back(make_pair(a[i], i));
      ba[i] = b[i];
      bd[i] = 2 * n + 1;
      ;
    } else {
      td++;
      ba[i] = 0;
      ad.push_back(make_pair(a[i], i));
      bd[i] = b[i];
    }
  }
  t = max(ta, td);
  int ans[t];
  if (ta > td) {
    ans[0] = max_element(ba, ba + n) - ba;
    sort(aa.begin(), aa.end(), comp);
    for (int i = 0; i < ta; i++) {
      if (aa[i].second != ans[0]) {
        ans[i] = aa[i].second;
      }
    }
  } else {
    ans[0] = min_element(bd, bd + n) - bd;
    sort(ad.begin(), ad.end(), compd1);
    for (int i = 0; i < td; i++) {
      if (ad[i].second != ans[0]) {
        ans[i] = ad[i].second;
      }
    }
  }
  printf("%d\n", t);
  for (int i = 0; i < t; i++) {
    printf("%d ", ans[i] + 1);
  }
}