#include <bits/stdc++.h>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
int a[300010], b[300010];
std::vector<std::pair<std::pair<int, int>, int>> vec1, vec2;
int main() {
  int n = inp();
  for (int i = 1; i <= n; i++) {
    a[i] = inp();
    b[i] = inp();
    if (a[i] < b[i])
      vec1.push_back(std::make_pair(std::make_pair(b[i], a[i]), i));
    else
      vec2.push_back(std::make_pair(std::make_pair(b[i], a[i]), i));
  }
  std::sort(vec1.begin(), vec1.end(),
            std::greater<std::pair<std::pair<int, int>, int>>());
  std::sort(vec2.begin(), vec2.end());
  if (vec1.size() > vec2.size()) {
    printf("%d\n", vec1.size());
    for (int i = 0; i < vec1.size(); i++) printf("%d ", vec1[i].second);
    putchar('\n');
  } else {
    printf("%d\n", vec2.size());
    for (int i = 0; i < vec2.size(); i++) printf("%d ", vec2[i].second);
    putchar('\n');
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
int inp() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum;
}
int a[300010], b[300010];
std::vector<std::pair<std::pair<int, int>, int>> vec1, vec2;
int main() {
  int n = inp();
  for (int i = 1; i <= n; i++) {
    a[i] = inp();
    b[i] = inp();
    if (a[i] < b[i])
      vec1.push_back(std::make_pair(std::make_pair(b[i], a[i]), i));
    else
      vec2.push_back(std::make_pair(std::make_pair(b[i], a[i]), i));
  }
  std::sort(vec1.begin(), vec1.end(),
            std::greater<std::pair<std::pair<int, int>, int>>());
  std::sort(vec2.begin(), vec2.end());
  if (vec1.size() > vec2.size()) {
    printf("%d\n", vec1.size());
    for (int i = 0; i < vec1.size(); i++) printf("%d ", vec1[i].second);
    putchar('\n');
  } else {
    printf("%d\n", vec2.size());
    for (int i = 0; i < vec2.size(); i++) printf("%d ", vec2[i].second);
    putchar('\n');
  }
}

```