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;
const int maxn = 300007;
struct tup {
  int u, v, pos;
  friend bool operator<(tup x, tup y) {
    if (x.v < y.v) {
      return 1;
    }
    return 0;
  }
} ans[maxn];
int rec[maxn], n;
int solve() {
  int cnt = 0;
  rec[cnt++] = n;
  ans[n].v = 2 * n + 1;
  for (int i = n - 1; i >= 0; --i) {
    if (ans[i].u < ans[rec[cnt - 1]].v && ans[i].u < ans[i].v) {
      rec[cnt++] = i;
    }
  }
  return cnt;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &ans[i].u, &ans[i].v);
    ans[i].pos = i;
  }
  sort(ans, ans + n);
  int cnt1 = solve() - 1;
  int tmp[maxn];
  for (int i = 1; i <= cnt1; ++i) {
    tmp[i] = ans[rec[i]].pos + 1;
  }
  for (int i = 0; i < n; ++i) {
    ans[i].u = -ans[i].u;
    ans[i].v = -ans[i].v;
  }
  sort(ans, ans + n);
  int cnt = solve() - 1;
  if (cnt <= cnt1) {
    printf("%d\n", cnt1);
    for (int i = 1; i <= cnt1; ++i) {
      printf("%d ", tmp[i]);
    }
  } else {
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; ++i) {
      printf("%d ", ans[rec[i]].pos + 1);
    }
  }
  return 0;
}