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;
void read(int &x) {
  char ch;
  bool ok;
  for (ok = 0, ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') ok = 1;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  if (ok) x = -x;
}
const int maxn = 3e5 + 10;
struct oo {
  int a, b, id;
} f[maxn], g[maxn];
int n, tot, sum;
bool cmp1(oo a, oo b) { return a.b > b.b; }
bool cmp2(oo a, oo b) { return a.a < b.a; }
int main() {
  read(n);
  for (register int i = 1, x, y; i <= n; i++) {
    read(x), read(y);
    if (y > x)
      f[++tot] = (oo){x, y, i};
    else
      g[++sum] = (oo){x, y, i};
  }
  if (tot > sum) {
    sort(f + 1, f + tot + 1, cmp1);
    printf("%d\n", tot);
    for (register int i = 1; i <= tot; i++)
      printf("%d%c", f[i].id, i == tot ? '\n' : ' ');
  } else {
    sort(g + 1, g + sum + 1, cmp2);
    printf("%d\n", sum);
    for (register int i = 1; i <= sum; i++)
      printf("%d%c", g[i].id, i == sum ? '\n' : ' ');
  }
}