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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
template <class T>
inline T bigMod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline string int2String(T a) {
  ostringstream str;
  str << a;
  return str.str();
}
template <class T>
inline T string2int(string a) {
  stringstream str(a);
  T b;
  str >> b;
  return b;
}
template <class T>
inline void read(T& r) {
  r = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' or ch > '9') {
    f = (ch == '-' ? -1 : 1), ch = getchar();
  }
  while (ch >= '0' and ch <= '9') {
    r = r * 10 + ch - '0', ch = getchar();
  }
  r *= f;
}
const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1, -2, -2, 2, 2, -1, -1, 1, 1};
const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1, -1, 1, -1, 1, -2, 2, -2, 2};
vector<int> a, b, ans;
pair<int, int> tmp[1 << 20];
bool cmp(int a, int b) { return tmp[a] < tmp[b]; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = int(0); i < int(n); i++) {
    scanf("%d %d", &tmp[i].first, &tmp[i].second);
    if (tmp[i].first > tmp[i].second)
      a.push_back(i);
    else if (tmp[i].first < tmp[i].second)
      b.push_back(i);
  }
  sort(a.begin(), a.end(), cmp);
  sort(b.begin(), b.end(), cmp);
  if (int(a.size()) > int(b.size())) {
    ans = a;
  } else {
    ans = b;
    reverse(ans.begin(), ans.end());
  }
  cout << int(ans.size()) << endl;
  for (auto x : ans) cout << x + 1 << " ";
  cout << endl;
  return 0;
}