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;
inline int read() {
  int x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
const int N = 3e5 + 10, INF = 0x3f3f3f3f;
struct Data {
  pair<int, int> v;
  int id;
} a[N];
inline bool cmp1(const Data& t1, const Data& t2) {
  return t1.v.second > t2.v.second;
}
inline bool cmp2(const Data& t1, const Data& t2) {
  return t1.v.second < t2.v.second;
}
int tmp[N], lt;
int tans[N], lans;
int main() {
  int n = read();
  for (int i = 1; i <= n; ++i)
    a[i].v.first = read(), a[i].v.second = read(), a[i].id = i;
  sort(a + 1, a + n + 1, cmp1);
  int now = INF;
  for (int i = 1; i <= n; ++i) {
    if (a[i].v.first > a[i].v.second) continue;
    if (a[i].v.first < now) {
      now = a[i].v.second;
      tmp[++lt] = a[i].id;
    }
  }
  lans = lt, memcpy(tans, tmp, sizeof(tmp));
  sort(a + 1, a + n + 1, cmp2);
  now = lt = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i].v.first < a[i].v.second) continue;
    if (a[i].v.first > now) {
      now = a[i].v.second;
      tmp[++lt] = a[i].id;
    }
  }
  if (lt > lans) {
    lans = lt, memcpy(tans, tmp, sizeof(tmp));
  }
  printf("%d\n", lans);
  for (int i = 1; i <= lans; ++i)
    printf("%d%c", tans[i], i == lans ? '\n' : ' ');
  return 0;
}