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;
int mod = 998244353;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T>
T GCD(T a, T b) {
  long long t;
  while (a) {
    t = a;
    a = b % a;
    b = t;
  }
  return b;
}
template <typename T>
string toString(T a) {
  return to_string(a);
}
template <typename T>
void toInt(string s, T& x) {
  stringstream str(s);
  str >> x;
}
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
inline int powr(int a, long long b) {
  int x = 1 % mod;
  while (b) {
    if (b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a) { return powr(a, mod - 2); }
bool comp1(pair<int, int> p1, pair<int, int> p2) {
  return p1.second > p2.second;
}
bool comp2(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<pair<int, int> > vec1, vec2;
  int a, b;
  map<pair<int, int>, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    if (a < b)
      vec1.push_back({a, b});
    else
      vec2.push_back({a, b});
    mp[{a, b}] = i + 1;
  }
  int ans1 = 0, ans2 = 0;
  vector<int> c1, c2;
  sort(vec1.begin(), vec1.end(), comp1);
  sort(vec2.begin(), vec2.end(), comp2);
  int cur = 2 * n + 1;
  for (auto it : vec1) {
    if (it.first < cur) {
      c1.push_back(mp[it]);
      cur = it.second;
    }
  }
  cur = 0;
  for (auto it : vec2) {
    if (it.first > cur) {
      c2.push_back(mp[it]);
      cur = it.second;
    }
  }
  if (c1.size() >= c2.size()) {
    cout << c1.size() << '\n';
    for (auto it : c1) {
      cout << it << " ";
    }
    cout << '\n';
  } else {
    cout << c2.size() << '\n';
    for (auto it : c2) {
      cout << it << " ";
    }
    cout << '\n';
  }
  return 0;
}