#include <bits/stdc++.h>
using namespace std;
const int N = 2e6;
const long long INF = 1e10;
struct elem {
  int x, y, id;
  elem() {}
} a[N];
bool cmp_x(const elem &v, const elem &u) {
  return pair<int, int>(v.x, v.y) < pair<int, int>(u.x, u.y);
}
bool cmp_ly(const elem &v, const elem &u) {
  return pair<int, int>(v.y, v.x) < pair<int, int>(u.y, u.x);
}
bool cmp_gy(const elem &v, const elem &u) {
  return pair<int, int>(v.y, v.x) > pair<int, int>(u.y, u.x);
}
int operator-(const elem &v, const elem &u) {
  return abs(v.x - u.x) + abs(v.y - u.y);
}
const int block = 1000;
int n;
vector<elem> v[2 * block];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = int(1); i <= int(n); ++i) {
    a[i].id = i;
    cin >> a[i].x >> a[i].y;
    v[a[i].x / block].push_back(a[i]);
  }
  for (int i = int(0); i <= int(block); ++i)
    if (i & 1)
      sort(v[i].begin(), v[i].end(), cmp_ly);
    else
      sort(v[i].begin(), v[i].end(), cmp_gy);
  long long ans = 0;
  int last = 0;
  for (int i = int(0); i <= int(block); ++i) {
    for (auto &x : v[i]) {
      cout << x.id << " ";
      if (last) ans += x - a[last];
      last = x.id;
    }
  }
  assert(ans <= 25 * 1e8);
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

On a plane are n points (xi, yi) with integer coordinates between 0 and 106. The distance between the two points with numbers a and b is said to be the following value: <image> (the distance calculated by such formula is called Manhattan distance).

We call a hamiltonian path to be some permutation pi of numbers from 1 to n. We say that the length of this path is value <image>.

Find some hamiltonian path with a length of no more than 25 × 108. Note that you do not have to minimize the path length.

Input

The first line contains integer n (1 ≤ n ≤ 106).

The i + 1-th line contains the coordinates of the i-th point: xi and yi (0 ≤ xi, yi ≤ 106).

It is guaranteed that no two points coincide.

Output

Print the permutation of numbers pi from 1 to n — the sought Hamiltonian path. The permutation must meet the inequality <image>.

If there are multiple possible answers, print any of them.

It is guaranteed that the answer exists.

Examples

Input

5
0 7
8 10
3 4
5 0
9 12


Output

4 3 1 2 5 

Note

In the sample test the total distance is:

<image>

(|5 - 3| + |0 - 4|) + (|3 - 0| + |4 - 7|) + (|0 - 8| + |7 - 10|) + (|8 - 9| + |10 - 12|) = 2 + 4 + 3 + 3 + 8 + 3 + 1 + 2 = 26

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6;
const long long INF = 1e10;
struct elem {
  int x, y, id;
  elem() {}
} a[N];
bool cmp_x(const elem &v, const elem &u) {
  return pair<int, int>(v.x, v.y) < pair<int, int>(u.x, u.y);
}
bool cmp_ly(const elem &v, const elem &u) {
  return pair<int, int>(v.y, v.x) < pair<int, int>(u.y, u.x);
}
bool cmp_gy(const elem &v, const elem &u) {
  return pair<int, int>(v.y, v.x) > pair<int, int>(u.y, u.x);
}
int operator-(const elem &v, const elem &u) {
  return abs(v.x - u.x) + abs(v.y - u.y);
}
const int block = 1000;
int n;
vector<elem> v[2 * block];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = int(1); i <= int(n); ++i) {
    a[i].id = i;
    cin >> a[i].x >> a[i].y;
    v[a[i].x / block].push_back(a[i]);
  }
  for (int i = int(0); i <= int(block); ++i)
    if (i & 1)
      sort(v[i].begin(), v[i].end(), cmp_ly);
    else
      sort(v[i].begin(), v[i].end(), cmp_gy);
  long long ans = 0;
  int last = 0;
  for (int i = int(0); i <= int(block); ++i) {
    for (auto &x : v[i]) {
      cout << x.id << " ";
      if (last) ans += x - a[last];
      last = x.id;
    }
  }
  assert(ans <= 25 * 1e8);
  return 0;
}

```