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;
template <typename T>
inline int Chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline int Chkmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  T sum = 0, fl = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fl = -1;
  for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
  return sum * fl;
}
inline void proc_status() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
const int Maxn = 1e6 + 100;
int N, Ra[Maxn], Rb[Maxn];
int Rank[Maxn];
vector<pair<pair<int, int>, int> > A, B;
inline int cmp1(int a, int b) { return A[a].first.second > A[b].first.second; }
inline int cmp2(int a, int b) { return B[a].first.first < B[b].first.first; }
inline void Solve() {
  int cnt0 = 0, cnt1 = 0;
  for (int i = 1; i <= N; ++i) {
    if (Ra[i] < Rb[i])
      A.push_back(make_pair(make_pair(Ra[i], Rb[i]), i));
    else
      B.push_back(make_pair(make_pair(Ra[i], Rb[i]), i));
  }
  for (int i = 0; i < N; ++i) Rank[i] = i;
  if (A.size() >= B.size()) {
    int n = A.size();
    cout << n << endl;
    sort(Rank, Rank + n, cmp1);
    for (int i = 0; i < n; ++i) cout << A[Rank[i]].second << ' ';
  } else {
    int n = B.size();
    cout << n << endl;
    sort(Rank, Rank + n, cmp2);
    for (int i = 0; i < n; ++i) cout << B[Rank[i]].second << ' ';
  }
}
inline void Input() {
  N = read<int>();
  for (int i = 1; i <= N; ++i) Ra[i] = read<int>(), Rb[i] = read<int>();
}
int main() {
  Input();
  Solve();
  return 0;
}