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 double pi = acos(-1);
const double eps = 1e-6;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 100;
int cnt1, cnt2, ans1[maxn], ans2[maxn], tot1, tot2;
struct node {
  int a, b, num;
  bool operator<(const node& z) const { return b > z.b; }
} x[maxn];
struct node2 {
  int a, b, num;
  bool operator<(const node2& z) const { return b < z.b; }
} y[maxn];
void debug() {}
void solve() {
  int n;
  scanf("%d", &n);
  int a, b;
  for (int i = 0; (i) < (n); i++) {
    scanf("%d%d", &a, &b);
    if (a < b) x[cnt1].num = i + 1, x[cnt1].a = a, x[cnt1++].b = b;
    if (a > b) y[cnt2].num = i + 1, y[cnt2].a = a, y[cnt2++].b = b;
  }
  sort(x, x + cnt1);
  sort(y, y + cnt2);
  int now = x[0].b;
  if (cnt1) ans1[tot1++] = x[0].num;
  for (int i = 1; (i) <= (cnt1 - 1); i++) {
    if (x[i].a < now) {
      ans1[tot1++] = x[i].num;
      now = x[i].b;
    }
  }
  now = y[0].b;
  if (cnt2) ans2[tot2++] = y[0].num;
  for (int i = 1; (i) <= (cnt2 - 1); i++) {
    if (y[i].a > now) {
      ans2[tot2++] = y[i].num;
      now = y[i].b;
    }
  }
  if (tot1 > tot2) {
    printf("%d\n", tot1);
    for (int i = 0; (i) < (tot1); i++) printf("%d ", ans1[i]);
  } else {
    printf("%d\n", tot2);
    for (int i = 0; (i) < (tot2); i++) printf("%d ", ans2[i]);
  }
}
int main() {
  solve();
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  return 0;
}