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;
int n, a[300009], b[300009], pos[600009];
void mergesort(int l, int r) {
  if (r - l + 1 < 2) return;
  if (r - l + 1 == 2) {
    if (a[l] > a[r]) swap(a[r], a[l]);
    return;
  } else {
    int mid = (l + r) / 2, start1 = l, start2 = mid + 1;
    mergesort(l, mid);
    mergesort(mid + 1, r);
    int ll = 0, *p = NULL;
    p = new int[r - l + 1];
    while (start1 <= mid && start2 <= r) {
      if (a[start1] < a[start2]) {
        p[ll] = a[start1];
        start1++;
      } else {
        p[ll] = a[start2];
        start2++;
      }
      ll++;
    }
    while (start1 <= mid) {
      p[ll] = a[start1];
      start1++;
      ll++;
    }
    while (start2 <= r) {
      p[ll] = a[start2];
      start2++;
      ll++;
    }
    for (int i = 0; i < ll; i++) {
      a[l + i] = p[i];
    }
    free(p);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  int c1 = 0, c2 = 0;
  for (int i = 1; i <= n; i++) {
    int u, v;
    cin >> u >> v;
    if (u > v) {
      c1++;
      b[c1] = u;
      pos[b[c1]] = i;
    } else {
      c2++;
      a[c2] = u;
      pos[a[c2]] = i;
    }
  }
  if (c2 > c1) {
    mergesort(1, c2);
    cout << c2 << "\n";
    for (int i = c2; i >= 1; i--) {
      cout << pos[a[i]] << " ";
    }
  } else {
    for (int i = 1; i <= c1; i++) a[i] = b[i];
    cout << c1 << "\n";
    mergesort(1, c1);
    for (int i = 1; i <= c1; i++) {
      cout << pos[a[i]] << " ";
    }
  }
  return 0;
}