#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000006;
struct node {
  int x, y, pos;
} a[1000100];
vector<pair<int, int>> _a[1010];
bool cmp(const pair<int, int> aa, const pair<int, int> bb) {
  return aa.first < bb.first;
}
int main() {
  int n, k, x;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i <= 1000; i++) _a[i].clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &a[i].x, &a[i].y);
      int d;
      _a[a[i].x / 1000].push_back(pair<int, int>(a[i].y, i));
    }
    int p = 1;
    for (int i = 0; i <= 1000; i++) {
      if (_a[i].empty()) continue;
      sort(_a[i].begin(), _a[i].end(), cmp);
      p ^= 1;
      if (p)
        for (vector<pair<int, int>>::reverse_iterator rit = _a[i].rbegin();
             rit != _a[i].rend(); rit++)
          printf("%d ", rit->second);
      else
        for (vector<pair<int, int>>::iterator it = _a[i].begin();
             it != _a[i].end(); it++)
          printf("%d ", it->second);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 maxn = 2000006;
struct node {
  int x, y, pos;
} a[1000100];
vector<pair<int, int>> _a[1010];
bool cmp(const pair<int, int> aa, const pair<int, int> bb) {
  return aa.first < bb.first;
}
int main() {
  int n, k, x;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i <= 1000; i++) _a[i].clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d %d", &a[i].x, &a[i].y);
      int d;
      _a[a[i].x / 1000].push_back(pair<int, int>(a[i].y, i));
    }
    int p = 1;
    for (int i = 0; i <= 1000; i++) {
      if (_a[i].empty()) continue;
      sort(_a[i].begin(), _a[i].end(), cmp);
      p ^= 1;
      if (p)
        for (vector<pair<int, int>>::reverse_iterator rit = _a[i].rbegin();
             rit != _a[i].rend(); rit++)
          printf("%d ", rit->second);
      else
        for (vector<pair<int, int>>::iterator it = _a[i].begin();
             it != _a[i].end(); it++)
          printf("%d ", it->second);
    }
    printf("\n");
  }
  return 0;
}

```