#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vii = vector<pii>;
using vs = vector<string>;
const int DR[] = {-1, 0, 1, 0};
const int DC[] = {0, -1, 0, 1};
const double PI = acos(-1.0);
const double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int INF = 1073741823;
const ll INFLL = 4e18;
const int MAX = 1e3;
int N;
vi grid[MAX + 5][MAX + 5];
void read() {
  cin >> N;
  for (int i = 0; i < (int)N; i++) {
    int r, c;
    cin >> c >> r;
    grid[r / 1000][c / 1000].push_back(i + 1);
  }
}
void solve() {
  for (int i = (int)0; i <= (int)1000; i++) {
    if (i % 2 == 0) {
      for (int j = 0; j <= 1000; j++)
        for (int x : grid[i][j]) cout << x << " ";
    } else {
      for (int j = 1000; j >= 0; j--)
        for (int x : grid[i][j]) cout << x << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TC = 1;
  while (TC--) {
    read();
    solve();
  }
}

### Prompt

Your challenge is to write a Cpp solution 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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vi = vector<int>;
using vii = vector<pii>;
using vs = vector<string>;
const int DR[] = {-1, 0, 1, 0};
const int DC[] = {0, -1, 0, 1};
const double PI = acos(-1.0);
const double EPS = 1e-9;
const int MOD = 1e9 + 7;
const int INF = 1073741823;
const ll INFLL = 4e18;
const int MAX = 1e3;
int N;
vi grid[MAX + 5][MAX + 5];
void read() {
  cin >> N;
  for (int i = 0; i < (int)N; i++) {
    int r, c;
    cin >> c >> r;
    grid[r / 1000][c / 1000].push_back(i + 1);
  }
}
void solve() {
  for (int i = (int)0; i <= (int)1000; i++) {
    if (i % 2 == 0) {
      for (int j = 0; j <= 1000; j++)
        for (int x : grid[i][j]) cout << x << " ";
    } else {
      for (int j = 1000; j >= 0; j--)
        for (int x : grid[i][j]) cout << x << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TC = 1;
  while (TC--) {
    read();
    solve();
  }
}

```