#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const ld pi = acos(0) * 2;
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
int n;
int a[300001];
int b[300001];
vector<int> ab, ba;
int main() {
  read(n);
  for (int i = (1); i <= (n); i++) {
    read(a[i]);
    read(b[i]);
  }
  int cnt = 0;
  for (int i = (1); i <= (n); i++)
    if (a[i] < b[i])
      ab.push_back(i);
    else
      ba.push_back(i);
  writeln(max(ab.size(), ba.size()));
  if (ab.size() > ba.size()) {
    sort(ab.begin(), ab.end(), [](int x, int y) { return a[x] > a[y]; });
    for (int x : ab) {
      write(x);
      putchar(' ');
    }
  } else {
    sort(ba.begin(), ba.end(), [](int x, int y) { return a[x] < a[y]; });
    for (int x : ba) {
      write(x);
      putchar(' ');
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const ld pi = acos(0) * 2;
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
int n;
int a[300001];
int b[300001];
vector<int> ab, ba;
int main() {
  read(n);
  for (int i = (1); i <= (n); i++) {
    read(a[i]);
    read(b[i]);
  }
  int cnt = 0;
  for (int i = (1); i <= (n); i++)
    if (a[i] < b[i])
      ab.push_back(i);
    else
      ba.push_back(i);
  writeln(max(ab.size(), ba.size()));
  if (ab.size() > ba.size()) {
    sort(ab.begin(), ab.end(), [](int x, int y) { return a[x] > a[y]; });
    for (int x : ab) {
      write(x);
      putchar(' ');
    }
  } else {
    sort(ba.begin(), ba.end(), [](int x, int y) { return a[x] < a[y]; });
    for (int x : ba) {
      write(x);
      putchar(' ');
    }
  }
}

```