#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 7;
const int B = (int)1e3;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) {
    x = _x;
    y = _y;
  }
};
point a[N];
vector<int> b[B + 7];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    b[a[i].y / B].push_back(i);
  }
  for (int i = 0; i <= B; i++) {
    vector<int> res;
    for (int j = 0; j < b[i].size(); j++) {
      res.push_back(a[b[i][j]].x);
    }
    sort(res.begin(), res.end());
    res.resize(unique(res.begin(), res.end()) - res.begin());
    for (int j = 0; j < b[i].size(); j++) {
      a[b[i][j]].x =
          lower_bound(res.begin(), res.end(), a[b[i][j]].x) - res.begin();
    }
    if (~i & 1) {
      sort(b[i].begin(), b[i].end(), [&](int p, int q) {
        if (a[p].x != a[q].x) {
          return a[p].x < a[q].x;
        }
        if (~a[p].x & 1) {
          return a[p].y < a[q].y;
        } else {
          return a[p].y > a[q].y;
        }
      });
    } else {
      sort(b[i].begin(), b[i].end(), [&](int p, int q) {
        if (a[p].x != a[q].x) {
          return a[p].x > a[q].x;
        }
        if (~a[p].x & 1) {
          return a[p].y < a[q].y;
        } else {
          return a[p].y > a[q].y;
        }
      });
    }
    for (int j = 0; j < b[i].size(); j++) {
      printf("%d ", b[i][j] + 1);
    }
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = (int)1e6 + 7;
const int B = (int)1e3;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) {
    x = _x;
    y = _y;
  }
};
point a[N];
vector<int> b[B + 7];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    b[a[i].y / B].push_back(i);
  }
  for (int i = 0; i <= B; i++) {
    vector<int> res;
    for (int j = 0; j < b[i].size(); j++) {
      res.push_back(a[b[i][j]].x);
    }
    sort(res.begin(), res.end());
    res.resize(unique(res.begin(), res.end()) - res.begin());
    for (int j = 0; j < b[i].size(); j++) {
      a[b[i][j]].x =
          lower_bound(res.begin(), res.end(), a[b[i][j]].x) - res.begin();
    }
    if (~i & 1) {
      sort(b[i].begin(), b[i].end(), [&](int p, int q) {
        if (a[p].x != a[q].x) {
          return a[p].x < a[q].x;
        }
        if (~a[p].x & 1) {
          return a[p].y < a[q].y;
        } else {
          return a[p].y > a[q].y;
        }
      });
    } else {
      sort(b[i].begin(), b[i].end(), [&](int p, int q) {
        if (a[p].x != a[q].x) {
          return a[p].x > a[q].x;
        }
        if (~a[p].x & 1) {
          return a[p].y < a[q].y;
        } else {
          return a[p].y > a[q].y;
        }
      });
    }
    for (int j = 0; j < b[i].size(); j++) {
      printf("%d ", b[i][j] + 1);
    }
  }
}

```