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>
bool local = false;
using namespace std;
template <class T>
string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
void print(T v) {
  cout << ((int)(v).size()) << "\n";
  for (auto x : v) cout << x << ' ';
  cout << "\n";
};
template <class T>
void print1(T v) {
  cout << ((int)(v).size()) << "\n";
  for (auto x : v) cout << x + 1 << ' ';
  cout << "\n";
};
template <class T1, class T2>
ostream &operator<<(ostream &o, pair<T1, T2> x) {
  return o << x.first << ' ' << x.second;
}
template <class T1, class T2>
istream &operator>>(istream &o, pair<T1, T2> &x) {
  return o >> x.first >> x.second;
}
template <class T1, class T2>
pair<T1, T2> operator+(pair<T1, T2> a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
template <class T1, class T2>
pair<T1, T2> operator-(pair<T1, T2> a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
  return a;
}
template <class T1, class T2>
void operator+=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first += b.first;
  a.second += b.second;
}
template <class T1, class T2>
void operator-=(pair<T1, T2> &a, pair<T1, T2> b) {
  a.first -= b.first;
  a.second -= b.second;
}
int nint() {
  int x;
  cin >> x;
  return x;
}
double getTime() { return clock() / (double)CLOCKS_PER_SEC; };
mt19937 rnd(0);
int rand(int n) { return rnd() % n; }
int rand(int l, int r) { return rnd() % (r - l + 1) + l; }
const int mod = 1000000000 + 7;
void initIO() {
  if (local) {
    freopen("input.txt", "r", stdin);
    srand((int)time(0));
    rnd.seed((int)time(0));
  } else {
  }
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  };
}
void solve();
int main() {
  initIO();
  int tc = 1;
  for (int tt = 0; tt < (tc); tt++) {
    solve();
  }
  if (local)
    cout << "\n"
         << "time = " << getTime() << "\n";
}
int n;
int m;
vector<int> v1, v2;
pair<int, int> val[300500];
bool f1(int a, int b) { return val[a].second > val[b].second; }
bool f2(int a, int b) { return val[a].second < val[b].second; }
void solve() {
  cin >> n;
  for (int i = 0; i < (n); i++) {
    pair<int, int> o;
    cin >> o;
    val[i] = o;
    if (o.first > o.second)
      v2.emplace_back(i);
    else
      v1.emplace_back(i);
  }
  sort(v1.begin(), v1.end(), f1);
  sort(v2.begin(), v2.end(), f2);
  if (((int)(v1).size()) > ((int)(v2).size()))
    print1(v1);
  else
    print1(v2);
}