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 double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 3e6 + 100;
const int maxm = 2e6 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
struct Node {
  int x, y;
  int id;
} node[maxn];
vector<Node> a, b;
bool cmp(Node a, Node b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    node[i].id = i;
    scanf("%d %d", &node[i].x, &node[i].y);
    if (node[i].x < node[i].y)
      a.push_back(node[i]);
    else
      b.push_back(node[i]);
  }
  printf("%d\n", max(a.size(), b.size()));
  sort(a.begin(), a.end(), cmp);
  sort(b.begin(), b.end(), cmp);
  if (a.size() > b.size()) {
    for (int i = a.size() - 1; i >= 0; i--) {
      printf("%d ", a[i].id);
    }
  } else {
    for (int i = 0; i < b.size(); i++) printf("%d ", b[i].id);
  }
  return 0;
}