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 N = 4 * 3e5 + 100;
pair<int, pair<int, int> > tree[4 * N];
int n;
void clear() {
  for (auto i = 0; i < (long long)(4 * N); i++) tree[i] = {-1, {-1, -1}};
}
void update(int node, int st, int en, int shemal, int yemeen,
            pair<int, int> val) {
  if (st > en || st > yemeen || en < shemal) return;
  if (shemal <= st && en <= yemeen) {
    tree[node] = {val.first, {val.second, st}};
    return;
  }
  int mid = (st + en) / 2;
  update(2 * node, st, mid, shemal, yemeen, val);
  update(2 * node + 1, mid + 1, en, shemal, yemeen, val);
  tree[node] = max(tree[2 * node], tree[2 * node + 1]);
}
pair<int, pair<int, int> > query(int node, int st, int en, int shemal,
                                 int yemeen) {
  if (st > en || st > yemeen || en < shemal) return {-1, {-1, -1}};
  if (shemal <= st && en <= yemeen) return tree[node];
  int mid = (st + en) / 2;
  pair<int, pair<int, int> > a = query(2 * node, st, mid, shemal, yemeen);
  pair<int, pair<int, int> > b =
      query(2 * node + 1, mid + 1, en, shemal, yemeen);
  return max(a, b);
}
void modify(vector<pair<int, int> > &vec) {
  clear();
  int mn = 1e9;
  for (auto i = 0; i < (long long)(n); i++) {
    vec[i].first *= -1;
    vec[i].second *= -1;
    mn = min(mn, vec[i].first);
    mn = min(mn, vec[i].second);
  }
  mn = abs(mn);
  for (auto i = 0; i < (long long)(n); i++) {
    vec[i].first += mn, vec[i].second += mn;
    if (vec[i].first < vec[i].second)
      update(1, 0, N - 1, vec[i].first, vec[i].first, {vec[i].second, i + 1});
  }
}
vector<int> solve() {
  vector<int> ret;
  int last = N - 1;
  while (true) {
    pair<int, pair<int, int> > mx = query(1, 0, N - 1, 0, last);
    if (mx.first == -1) break;
    ret.push_back(mx.second.first);
    update(1, 0, N - 1, mx.second.second, mx.second.second, {-1, -1});
    last = mx.first;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  vector<pair<int, int> > vec(n);
  clear();
  for (auto i = 0; i < (long long)(n); i++) {
    int a, b;
    cin >> a >> b;
    if (a < b) update(1, 0, N - 1, a, a, {b, i + 1});
    vec[i] = {a, b};
  }
  vector<int> a, b;
  a = solve();
  modify(vec);
  b = solve();
  if (((int)((a).size())) < ((int)((b).size()))) a = b;
  cout << ((int)((a).size())) << '\n';
  for (auto i : a) cout << i << " ";
  return 0;
}