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;
#pragma warning(disable : 4996)
const int INF = 1000000000;
const int MOD = 998244353;
struct Node {
  int a, b, i;
} A[300000], B[300000];
int a = 0, b = 0;
bool cmp(Node& A, Node& B) { return A.a > B.a; }
bool cmp2(Node& A, Node& B) { return A.a < B.a; }
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    if (u < v)
      A[a++] = {u, v, i};
    else
      B[b++] = {u, v, i};
  }
  sort(A, A + a, cmp);
  sort(B, B + b, cmp2);
  if (a > b) {
    printf("%d\n", a);
    for (int i = 0; i < a; i++) {
      printf("%d ", A[i].i + 1);
    }
  } else {
    printf("%d\n", b);
    for (int i = 0; i < b; i++) {
      printf("%d ", B[i].i + 1);
    }
  }
  puts("");
  return 0;
}