#include <bits/stdc++.h>
using namespace std;
const int LEAF = 1 << 20;
int mx[LEAF + LEAF], mn[LEAF + LEAF];
int a[300001], b[300001];
int pos[600001];
int _min(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return min(a, b);
}
void upd_mx(int ix, int val) {
  mx[ix] = val;
  ix >>= 1;
  while (ix) {
    mx[ix] = max(mx[ix + ix], mx[ix + ix + 1]);
    ix >>= 1;
  }
}
void upd_mn(int ix, int val) {
  mn[ix] = val;
  ix >>= 1;
  while (ix) {
    mn[ix] = _min(mn[ix + ix], mn[ix + ix + 1]);
    ix >>= 1;
  }
}
int rmx(int ix, int s, int e, int from, int to) {
  if (to < s || e < from) return 0;
  if (from <= s && e <= to) return mx[ix];
  return max(rmx(ix + ix, s, (s + e) >> 1, from, to),
             rmx(ix + ix + 1, 1 + ((s + e) >> 1), e, from, to));
}
int rmn(int ix, int s, int e, int from, int to) {
  if (to < s || e < from) return 0;
  if (from <= s && e <= to) return mn[ix];
  return _min(rmn(ix + ix, s, (s + e) >> 1, from, to),
              rmn(ix + ix + 1, 1 + ((s + e) >> 1), e, from, to));
}
int main() {
  int i, n;
  cin >> n;
  vector<int> inc, dec;
  for (i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i];
    if (a[i] < b[i]) {
      upd_mx(a[i] + LEAF, b[i]);
    } else {
      upd_mn(a[i] + LEAF, b[i]);
    }
    pos[b[i]] = i;
  }
  int top = mx[1];
  while (top) {
    i = pos[top];
    inc.push_back(i);
    upd_mx(a[i] + LEAF, 0);
    top = rmx(1, 0, LEAF - 1, 1, top);
  }
  top = mn[1];
  while (top) {
    i = pos[top];
    dec.push_back(i);
    upd_mn(a[i] + LEAF, 0);
    top = rmn(1, 0, LEAF - 1, top, LEAF);
  }
  if (inc.size() < dec.size()) {
    cout << dec.size() << endl;
    for (auto x : dec) cout << x << " ";
  } else {
    cout << inc.size() << endl;
    for (auto x : inc) cout << x << " ";
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int LEAF = 1 << 20;
int mx[LEAF + LEAF], mn[LEAF + LEAF];
int a[300001], b[300001];
int pos[600001];
int _min(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return min(a, b);
}
void upd_mx(int ix, int val) {
  mx[ix] = val;
  ix >>= 1;
  while (ix) {
    mx[ix] = max(mx[ix + ix], mx[ix + ix + 1]);
    ix >>= 1;
  }
}
void upd_mn(int ix, int val) {
  mn[ix] = val;
  ix >>= 1;
  while (ix) {
    mn[ix] = _min(mn[ix + ix], mn[ix + ix + 1]);
    ix >>= 1;
  }
}
int rmx(int ix, int s, int e, int from, int to) {
  if (to < s || e < from) return 0;
  if (from <= s && e <= to) return mx[ix];
  return max(rmx(ix + ix, s, (s + e) >> 1, from, to),
             rmx(ix + ix + 1, 1 + ((s + e) >> 1), e, from, to));
}
int rmn(int ix, int s, int e, int from, int to) {
  if (to < s || e < from) return 0;
  if (from <= s && e <= to) return mn[ix];
  return _min(rmn(ix + ix, s, (s + e) >> 1, from, to),
              rmn(ix + ix + 1, 1 + ((s + e) >> 1), e, from, to));
}
int main() {
  int i, n;
  cin >> n;
  vector<int> inc, dec;
  for (i = 1; i <= n; ++i) {
    cin >> a[i] >> b[i];
    if (a[i] < b[i]) {
      upd_mx(a[i] + LEAF, b[i]);
    } else {
      upd_mn(a[i] + LEAF, b[i]);
    }
    pos[b[i]] = i;
  }
  int top = mx[1];
  while (top) {
    i = pos[top];
    inc.push_back(i);
    upd_mx(a[i] + LEAF, 0);
    top = rmx(1, 0, LEAF - 1, 1, top);
  }
  top = mn[1];
  while (top) {
    i = pos[top];
    dec.push_back(i);
    upd_mn(a[i] + LEAF, 0);
    top = rmn(1, 0, LEAF - 1, top, LEAF);
  }
  if (inc.size() < dec.size()) {
    cout << dec.size() << endl;
    for (auto x : dec) cout << x << " ";
  } else {
    cout << inc.size() << endl;
    for (auto x : inc) cout << x << " ";
  }
  return 0;
}

```