#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % MOD;
    a = a * a % MOD;
    b /= 2;
  }
  return ans;
}
const int N = 1e6 + 11;
int n;
struct uzi {
  int a;
  int b;
  int i;
} p[N];
vector<int> res, cnt;
bool cmp1(const uzi &A, const uzi &B) { return A.b > B.b; }
bool cmp2(const uzi &A, const uzi &B) { return A.b < B.b; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].a >> p[i].b, p[i].i = i;
  sort(p + 1, p + 1 + n, cmp1);
  int las = 2 * n + 3;
  for (int i = 1; i <= n; i++)
    if (p[i].a < p[i].b && p[i].a < las) res.push_back(p[i].i), las = p[i].b;
  sort(p + 1, p + 1 + n, cmp2);
  las = 0;
  for (int i = 1; i <= n; i++)
    if (p[i].a > p[i].b && p[i].a > las) cnt.push_back(p[i].i), las = p[i].b;
  int A = (int)res.size(), B = (int)cnt.size();
  if (A >= B) {
    cout << A << endl;
    for (auto k : res) cout << k << ' ';
  } else {
    cout << B << endl;
    for (auto k : cnt) cout << k << ' ';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % MOD;
    a = a * a % MOD;
    b /= 2;
  }
  return ans;
}
const int N = 1e6 + 11;
int n;
struct uzi {
  int a;
  int b;
  int i;
} p[N];
vector<int> res, cnt;
bool cmp1(const uzi &A, const uzi &B) { return A.b > B.b; }
bool cmp2(const uzi &A, const uzi &B) { return A.b < B.b; }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].a >> p[i].b, p[i].i = i;
  sort(p + 1, p + 1 + n, cmp1);
  int las = 2 * n + 3;
  for (int i = 1; i <= n; i++)
    if (p[i].a < p[i].b && p[i].a < las) res.push_back(p[i].i), las = p[i].b;
  sort(p + 1, p + 1 + n, cmp2);
  las = 0;
  for (int i = 1; i <= n; i++)
    if (p[i].a > p[i].b && p[i].a > las) cnt.push_back(p[i].i), las = p[i].b;
  int A = (int)res.size(), B = (int)cnt.size();
  if (A >= B) {
    cout << A << endl;
    for (auto k : res) cout << k << ' ';
  } else {
    cout << B << endl;
    for (auto k : cnt) cout << k << ' ';
  }
  return 0;
}

```