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 = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 3e5 + 10;
int n, cnt1, cnt2, lo1, lo2, ha1[N], ha2[N];
struct Node {
  int a, b, c;
} a[N];
inline bool cmp1(Node x, Node y) { return x.a > y.a; }
inline bool cmp2(Node x, Node y) { return x.a < y.a; }
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    a[i].a = read();
    a[i].b = read();
    a[i].c = i;
  }
  sort(a + 1, a + n + 1, cmp1);
  for (int i = 1; i <= n; ++i)
    if (a[i].a < a[i].b) ha1[++cnt1] = a[i].c;
  sort(a + 1, a + n + 1, cmp2);
  for (int i = 1; i <= n; ++i)
    if (a[i].a > a[i].b) ha2[++cnt2] = a[i].c;
  if (cnt1 > cnt2) {
    printf("%d\n", cnt1);
    for (int i = 1; i <= cnt1; ++i) printf("%d ", ha1[i]);
  } else {
    printf("%d\n", cnt2);
    for (int i = 1; i <= cnt2; ++i) printf("%d ", ha2[i]);
  }
  return 0;
}