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;
long long z = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % z;
      b--;
    }
    a = (a * a) % z;
    b = b >> 1;
  }
  return res;
}
vector<long long> sv;
long long fact[200001];
vector<long long> pp;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  pp.assign(1000001, 0);
  for (long long i = 2; i < 1000; i++) {
    if (!pp[i]) {
      for (long long j = i * i; j <= 1000000; j += i) {
        pp[j] = 1;
      }
    }
  }
  fact[0] = 1;
  for (long long i = 1; i < 200001; i++)
    fact[i] = i * fact[i - 1], fact[i] %= z;
  for (long long i = 2; i <= 1000000; i++)
    if (!pp[i]) sv.push_back(i);
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<pair<pair<long long, long long>, long long> > v1, v2;
    long long n1 = 0, n2 = 0;
    for (long long i = 0; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      if (x < y)
        v1.push_back(make_pair(make_pair(x, y), i + 1)), n1++;
      else
        v2.push_back(make_pair(make_pair(x, y), i + 1)), n2++;
    }
    sort((v1).begin(), (v1).end());
    sort((v2).begin(), (v2).end());
    vector<long long> ans1, ans2;
    if (n1) {
      long long dp1[n1], in1[n1], in11[n1];
      for (long long i = 0; i < n1; i++) dp1[i] = 1, in1[i] = i, in11[i] = i;
      long long mx1 = 0, val1 = 1;
      for (long long i = 1; i < n1; i++) {
        long long l = 0, r = i - 1, in = -1;
        while (l <= r) {
          long long mid = l + (r - l) / 2;
          if (v1[mid].first.first < v1[i].first.second) {
            in = mid;
            l = mid + 1;
          } else
            r = mid - 1;
        }
        if (in != -1) {
          long long p = in11[in];
          dp1[i] += dp1[p];
          in1[i] = p;
        }
        if (dp1[i] < dp1[in11[i - 1]]) {
          in11[i] = in11[i - 1];
        }
        if (dp1[i] > val1) {
          val1 = dp1[i];
          mx1 = i;
        }
      }
      long long i = mx1;
      while (in1[i] != i) {
        ans1.push_back(v1[i].second);
        i = in1[i];
      }
      ans1.push_back(v1[i].second);
    }
    if (n2) {
      long long dp2[n2], in2[n2], in22[n2];
      for (long long i = 0; i < n2; i++) dp2[i] = 1, in2[i] = i, in22[i] = i;
      long long mx2 = n2 - 1, val2 = 1;
      for (long long i = n2 - 2; i >= 0; i--) {
        long long l = i + 1, r = n2 - 1, in = -1;
        while (l <= r) {
          long long mid = l + (r - l) / 2;
          if (v2[mid].first.first > v2[i].first.second) {
            in = mid;
            r = mid - 1;
          } else
            l = mid + 1;
        }
        if (in != -1) {
          long long p = in22[in];
          dp2[i] += dp2[p];
          in2[i] = p;
        }
        if (dp2[i] < dp2[in22[i + 1]]) {
          in22[i] = in22[i + 1];
        }
        if (dp2[i] > val2) {
          val2 = dp2[i];
          mx2 = i;
        }
      }
      long long i = mx2;
      while (in2[i] != i) {
        ans2.push_back(v2[i].second);
        i = in2[i];
      }
      ans2.push_back(v2[i].second);
    }
    vector<long long> ans = ans1.size() > ans2.size() ? ans1 : ans2;
    cout << ans.size() << "\n";
    for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << "\n";
  }
}