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 PRIME = 1e9 + 7;
const int INF = 2e9;
const int MAX_N = 1e3 + 6;
long long ans = 0;
int n, m = 0, k, r = 0;
struct parR {
  int a, b, id;
  bool operator<(const parR& p) const {
    if (b != p.b)
      return b > p.b;
    else
      return a > p.a;
  }
};
struct parM {
  int a, b, id;
  bool operator<(const parM& p) const {
    if (b != p.b)
      return b < p.b;
    else
      return a < p.a;
  }
};
vector<parM> tabM;
vector<parR> tabR;
vector<int> resM;
vector<int> resR;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  int ia, ib;
  parM tM;
  parR tR;
  for (int i = (1); i <= (n); ++i) {
    cin >> ia >> ib;
    if (ia < ib) {
      tR.a = ia;
      tR.b = ib;
      tR.id = i;
      tabR.push_back(tR);
      ++r;
    } else {
      tM.a = ia;
      tM.b = ib;
      tM.id = i;
      tabM.push_back(tM);
      ++m;
    }
  }
  sort(tabM.begin(), tabM.end());
  sort(tabR.begin(), tabR.end());
  int prv = INF;
  for (int pos = (0); pos < (r); ++pos) {
    if (prv > tabR[pos].a) {
      prv = tabR[pos].b;
      resR.push_back(tabR[pos].id);
    }
  }
  prv = 0;
  for (int pos = (0); pos < (m); ++pos) {
    if (prv < tabM[pos].a) {
      prv = tabM[pos].b;
      resM.push_back(tabM[pos].id);
    }
  }
  if (resR.size() > resM.size()) {
    cout << resR.size() << "\n";
    for (int v : resR) cout << v << " ";
  } else {
    cout << resM.size() << "\n";
    for (int v : resM) cout << v << " ";
  }
  return 0;
}