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 mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const int INF = 1e9 + 7;
struct note {
  int a, pos;
} a[maxn], b[maxn];
bool cmp1(note t1, note t2) { return t1.a < t2.a; }
bool cmp2(note t1, note t2) { return t1.a > t2.a; }
int main() {
  int n;
  scanf("%d", &n);
  int x, y;
  int l1 = 0, l2 = 0;
  for (int q = 1; q <= n; q++) {
    scanf("%d%d", &x, &y);
    if (x > y) {
      a[l1++] = note{x, q};
    } else {
      b[l2++] = note{x, q};
    }
  }
  if (l1 > l2) {
    sort(a, a + l1, cmp1);
    printf("%d\n", l1);
    for (int q = 0; q < l1; q++) printf("%d ", a[q].pos);
  } else {
    sort(b, b + l2, cmp2);
    printf("%d\n", l2);
    for (int q = 0; q < l2; q++) printf("%d ", b[q].pos);
  }
  return 0;
}