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;
const int Maxn = 300005;
int ans, cnt1, cnt2, n, x[Maxn], y[Maxn];
bool del1[Maxn], del2[Maxn];
struct line {
  int x, y, id;
  bool operator<(const line &tmp) const { return y > tmp.y; }
} L1[Maxn], L2[Maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++)
    if (x[i] < y[i]) L1[++cnt1].x = x[i], L1[cnt1].y = y[i], L1[cnt1].id = i;
  sort(L1 + 1, L1 + 1 + cnt1);
  ans = cnt1;
  for (int i = 1; i < n; i++)
    if (L1[i].y < L1[i + 1].x) ans--, del1[i] = true;
  int tmp = 0;
  for (int i = 1; i <= n; i++)
    if (x[i] > y[i]) L2[++cnt2].x = y[i], L2[cnt2].y = x[i], L2[cnt2].id = i;
  sort(L2 + 1, L2 + 1 + cnt2);
  tmp = cnt2;
  for (int i = 1; i < n; i++)
    if (L2[i].y < L2[i + 1].x) tmp--, del2[i] = true;
  printf("%d\n", max(tmp, ans));
  if (tmp < ans) {
    for (int i = 1; i <= cnt1; i++)
      if (!del1[i]) printf("%d ", L1[i].id);
  } else {
    for (int i = cnt2; i >= 1; i--)
      if (!del2[i]) printf("%d ", L2[i].id);
  }
  return 0;
}