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 <class T>
constexpr T lcm(T m, T n) {
  return m / __gcd(m, n) * n;
}
template <typename I>
void join(ostream &ost, I s, I t, string d = " ") {
  for (auto i = s; i != t; ++i) {
    if (i != s) ost << d;
    ost << *i;
  }
  ost << endl;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &a : v) is >> a;
  return is;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
bool chmin(T &a, const U &b) {
  return (a > b ? a = b, true : false);
}
template <typename T, typename U>
bool chmax(T &a, const U &b) {
  return (a < b ? a = b, true : false);
}
template <typename T, size_t N, typename U>
void fill_array(T (&a)[N], const U &v) {
  fill((U *)a, (U *)(a + N), v);
}
template <typename T>
class SegmentTree {
 private:
  int size;
  vector<T> vec;
  T e;
  function<T(T, T)> f, upd;
  inline T aux(int x, int y, int i, int l, int r) {
    if (r <= x || y <= l)
      return e;
    else if (x <= l && r <= y)
      return vec[i];
    else
      return f(aux(x, y, i * 2 + 1, l, (l + r) / 2),
               aux(x, y, i * 2 + 2, (l + r) / 2, r));
  };

 public:
  SegmentTree(int n, const T &e, const function<T(T, T)> &f,
              const function<T(T, T)> &upd)
      : e(e), f(f), upd(upd) {
    size = 1;
    while (size < n) size *= 2;
    size = size * 2 - 1;
    vec = vector<T>(size, e);
  }
  inline void update(int i, const T &x) {
    int j = i + (size + 1) / 2 - 1;
    vec[j] = upd(vec[j], x);
    --j;
    while (j >= 0) {
      vec[j / 2] = f(vec[(j / 2) * 2 + 1], vec[(j / 2) * 2 + 2]);
      (j /= 2) -= 1;
    }
  }
  inline T get(int x, int y) { return aux(x, y, 0, 0, (size + 1) / 2); }
};
vector<int> solve1(int n, const vector<int> &a, const vector<int> &b) {
  vector<int> x, y;
  map<int, int> m;
  for (long long int i = (0); i < (n); ++i) {
    if (a[i] < b[i]) {
      x.push_back(a[i]);
      y.push_back(b[i]);
    }
    m[b[i]] = i;
  }
  ((void)0);
  ((void)0);
  if (x.size() == 0) return {};
  vector<int> ret;
  auto f = [](int i, int j) { return max(i, j); };
  SegmentTree<int> seg(2 * n + 1, 0, f, [](int i, int j) { return j; });
  for (long long int i = (0); i < ((int)x.size()); ++i) {
    seg.update(x[i], y[i]);
  }
  int c;
  {
    c = seg.get(0, 2 * n + 1);
    int i = m[c];
    int k = a[i];
    ((void)0);
    seg.update(k, 0);
    ret.push_back(i);
  }
  while (1) {
    c = seg.get(0, c);
    if (c == 0) break;
    int i = m[c];
    int k = a[i];
    seg.update(k, 0);
    ret.push_back(i);
  }
  return ret;
}
vector<int> solve2(int n, const vector<int> &a, const vector<int> &b) {
  vector<int> x, y;
  map<int, int> m;
  for (long long int i = (0); i < (n); ++i) {
    if (a[i] > b[i]) {
      x.push_back(a[i]);
      y.push_back(b[i]);
    }
    m[b[i]] = i;
  }
  ((void)0);
  ((void)0);
  if (x.size() == 0) return {};
  vector<int> ret;
  auto f = [](int i, int j) { return min(i, j); };
  SegmentTree<int> seg(2 * n + 1, INT_MAX, f, [](int i, int j) { return j; });
  for (long long int i = (0); i < ((int)x.size()); ++i) {
    seg.update(x[i], y[i]);
  }
  int c;
  {
    c = seg.get(0, 2 * n + 1);
    int i = m[c];
    int k = a[i];
    ((void)0);
    seg.update(k, INT_MAX);
    ret.push_back(i);
  }
  while (1) {
    c = seg.get(c, 2 * n + 1);
    if (c == INT_MAX) break;
    int i = m[c];
    int k = a[i];
    seg.update(k, INT_MAX);
    ret.push_back(i);
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  while (cin >> n) {
    vector<int> a(n), b(n);
    for (long long int i = (0); i < (n); ++i) cin >> a[i] >> b[i];
    auto ans1 = solve1(n, a, b);
    auto ans2 = solve2(n, a, b);
    for (auto &x : ans1) ++x;
    for (auto &x : ans2) ++x;
    if (ans1.size() >= ans2.size()) {
      cout << ans1.size() << endl;
      join(cout, (ans1).begin(), (ans1).end());
    } else {
      cout << ans2.size() << endl;
      join(cout, (ans2).begin(), (ans2).end());
    }
  }
  return 0;
}