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 n, tot = 0;
long long c1 = 0, c2 = 0;
struct node {
  long long l, r, id;
} s[500005];
bool cmp1(node x, node y) { return x.r > y.r; }
bool cmp2(node x, node y) { return x.l < y.l; }
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    s[i].id = i;
    cin >> s[i].l >> s[i].r;
    if (s[i].l < s[i].r)
      c1++;
    else
      c2++;
  }
  cout << max(c1, c2) << endl;
  if (c1 >= c2) {
    sort(s + 1, s + 1 + n, cmp1);
    for (long long i = 1; i <= n; i++) {
      if (s[i].l < s[i].r) {
        cout << s[i].id << ' ';
      }
    }
  } else {
    sort(s + 1, s + 1 + n, cmp2);
    for (long long i = 1; i <= n; i++) {
      if (s[i].l > s[i].r) {
        cout << s[i].id << ' ';
      }
    }
  }
  return 0;
}