#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
struct node {
  int x, y, id;
} g1[1010101], g2[1010101];
int tot1, tot2, n;
inline bool cmp1(node a, node b) { return a.x < b.x; }
inline bool cmp2(node a, node b) { return a.x > b.x; }
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    if (x > y) {
      g1[++tot1].x = x;
      g1[tot1].y = y;
      g1[tot1].id = i;
    } else {
      g2[++tot2].x = x;
      g2[tot2].y = y;
      g2[tot2].id = i;
    }
  }
  sort(g1 + 1, g1 + 1 + tot1, cmp1);
  sort(g2 + 1, g2 + 1 + tot2, cmp2);
  if (tot1 >= tot2) {
    cout << tot1 << endl;
    for (int i = 1; i <= tot1; i++) {
      printf("%d ", g1[i].id);
    }
  } else {
    cout << tot2 << endl;
    for (int i = 1; i <= tot2; i++) {
      printf("%d ", g2[i].id);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
struct node {
  int x, y, id;
} g1[1010101], g2[1010101];
int tot1, tot2, n;
inline bool cmp1(node a, node b) { return a.x < b.x; }
inline bool cmp2(node a, node b) { return a.x > b.x; }
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    if (x > y) {
      g1[++tot1].x = x;
      g1[tot1].y = y;
      g1[tot1].id = i;
    } else {
      g2[++tot2].x = x;
      g2[tot2].y = y;
      g2[tot2].id = i;
    }
  }
  sort(g1 + 1, g1 + 1 + tot1, cmp1);
  sort(g2 + 1, g2 + 1 + tot2, cmp2);
  if (tot1 >= tot2) {
    cout << tot1 << endl;
    for (int i = 1; i <= tot1; i++) {
      printf("%d ", g1[i].id);
    }
  } else {
    cout << tot2 << endl;
    for (int i = 1; i <= tot2; i++) {
      printf("%d ", g2[i].id);
    }
  }
  return 0;
}

```