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 d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long isPrime(long long n) {
  if (n == 0 || n == 1) return 0;
  if (n % 2 == 0) return 2;
  if (n % 3 == 0) return 3;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0) return i;
    if (n % (i + 2) == 0) return i + 2;
  }
  return n;
}
map<long long, long long> par, val;
void ini() {
  for (long long i = 0; i < 500010; i++) {
    par[i] = i;
  }
}
long long root(long long x) {
  while (par[x] != x) {
    par[x] = par[par[x]];
    x = par[x];
  }
  return x;
}
void uni(long long x, long long y) {
  x = root(x);
  y = root(y);
  if (val[x] <= val[y])
    par[y] = par[x];
  else
    par[x] = par[y];
}
vector<vector<pair<long long, long long> > > v(200010);
vector<long long> ans;
map<long long, long long> vis, dis;
long long bfs(long long n) {
  vis[1] = 1;
  dis[1] = 0;
  long long f = 0, i;
  queue<long long> q;
  q.push(1);
  long long ans = 0;
  while (q.size() > 0) {
    long long x = q.front();
    q.pop();
    for (i = 0; i < v[x].size(); i++) {
      if (!vis[v[x][i].first]) {
        dis[v[x][i].first] = dis[x] + v[x][i].second;
        ans = max(ans, dis[v[x][i].first]);
        q.push(v[x][i].first);
        vis[v[x][i].first] = 1;
      }
    }
  }
  return ans;
}
vector<long long> a(1000000);
vector<long long> spf(1000000);
void sieve() {
  long long i, j;
  bool prime[1000000];
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (i = 2; i * i <= 1000000; i++) {
    if (prime[i] == true) {
      for (j = i * i; j <= 100000; j += i) {
        prime[j] = false;
        if (spf[j] == 0) spf[j] = i;
      }
    }
  }
}
int gcdExtended(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverse(int b, int m) {
  int x, y;
  int g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, k, i, j, h;
  long long t;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  vector<pair<long long, pair<long long, long long> > > b(n);
  for (i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    b[i].first = a[i].second;
    b[i].second.first = a[i].first;
    b[i].second.second = i;
  }
  sort(b.rbegin(), b.rend());
  vector<long long> ans1, ans2;
  for (i = 0; i < n; i++) {
    if (b[i].second.first > b[i].first) continue;
    if (!ans1.size()) {
      ans1.push_back(b[i].second.second);
    } else {
      long long x = ans1.back();
      if (a[x].second > b[i].second.first) {
        ans1.push_back(b[i].second.second);
      }
    }
  }
  sort(b.begin(), b.end());
  for (i = 0; i < n; i++) {
    if (b[i].second.first < b[i].first) continue;
    if (!ans2.size()) {
      ans2.push_back(b[i].second.second);
    } else {
      long long x = ans2.back();
      if (a[x].second < b[i].second.first) {
        ans2.push_back(b[i].second.second);
      }
    }
  }
  if (ans1.size() >= ans2.size()) {
    cout << ans1.size() << endl;
    for (i = 0; i < ans1.size(); i++) cout << ans1[i] + 1 << " ";
    return 0;
  }
  cout << ans2.size() << endl;
  for (i = 0; i < ans2.size(); i++) cout << ans2[i] + 1 << " ";
}