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>
long long mod(long long x) {
  if (x < 0) return -x;
  return x;
}
long long expo(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y % 2 == 1) res = (res * x);
    y = y >> 1;
    x = x * x;
  }
  return res;
}
using namespace std;
const int MAXN = 500005;
vector<long long> v;
long long a[1000000], b[1000000], c[MAXN], d[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  long long n, t, m, k, x = 0, y = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    if (a[i] > b[i]) {
      x++;
    } else {
      y++;
    }
  }
  set<pair<long long, long long>> s;
  if (x > y) {
    k = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] > b[i]) {
        c[k] = a[i];
        d[k] = b[i];
        k++;
        pair<long long, long long> p = make_pair(d[k - 1], i);
        s.insert(p);
      }
    }
  } else {
    k = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] < b[i]) {
        c[k] = a[i];
        d[k] = b[i];
        k++;
        pair<long long, long long> p = make_pair(d[k - 1], i);
        s.insert(p);
      }
    }
  }
  for (set<pair<long long, long long>>::iterator it = s.begin(); it != s.end();
       it++) {
    pair<long long, long long> p = *it;
    v.push_back(p.second);
  }
  cout << int(v.size()) << endl;
  long long i;
  if (x > y) {
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i] + 1 << " ";
    }
    return 0;
  }
  for (long long i = v.size() - 1; i >= 0; i--) {
    cout << v[i] + 1 << " ";
  }
  return 0;
}