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 = 3e5 + 7;
struct pir {
  int a, b, idx;
  pir() {}
  pir(int a, int b, int idx) {
    this->a = a;
    this->b = b;
    this->idx = idx;
  }
};
bool cmp(pir lvalue, pir rvalue) { return lvalue.b > rvalue.b; }
bool cmp3(pir lvalue, pir rvalue) { return lvalue.b < rvalue.b; }
struct cmp2 {
  bool operator()(pir lvalue, pir rvalue) const { return lvalue.b > rvalue.b; }
};
vector<pir> v1, v2;
vector<pir> sm;
vector<int> solve1() {
  set<pir, cmp2> left;
  sm.clear();
  left.clear();
  for (pir p : v1) {
    sm.push_back(pir(p.a, p.b, p.idx));
    left.insert(p);
  }
  sort(sm.begin(), sm.end(), cmp);
  int i = 0;
  vector<int> ret;
  while (!left.empty()) {
    pir big = *left.begin();
    ret.push_back(big.idx);
    if (left.count(big)) {
      left.erase(big);
    }
    cout << big.a << " " << big.b << " " << big.idx << endl;
    while (i < (int)sm.size()) {
      cout << sm[i].a << " " << sm[i].b << endl;
      if (sm[i].a > big.b) {
        if (left.count(sm[i])) {
          left.erase(sm[i]);
        }
        i++;
      } else {
        break;
      }
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    cin >> a >> b;
    if (a < b) {
      v1.push_back(pir(a, b, i));
    } else {
      v2.push_back(pir(a, b, i));
    }
  }
  if (!v1.empty()) sort(v1.begin(), v1.end(), cmp);
  if (!v2.empty()) sort(v2.begin(), v2.end(), cmp3);
  if (v1.size() > v2.size()) {
    cout << (int)v1.size() << "\n";
    for (pir p : v1) {
      cout << p.idx << " ";
    }
    cout << "\n";
  } else {
    cout << (int)v2.size() << "\n";
    for (pir p : v2) {
      cout << p.idx << " ";
    }
    cout << "\n";
  }
  return 0;
}