#include <bits/stdc++.h>
using namespace std;
const int max_n = 1111111;
const int edge = 500;
const int sz = 2000;
int n;
int x[max_n], y[max_n];
vector<int> v[sz][sz];
bool flag;
int cmp(const int a, const int b) {
  if (x[a] != x[b]) return x[a] < x[b];
  return y[a] < y[b];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    int x_id = min(x[i] / edge, sz - 1), y_id = min(y[i] / edge, sz - 1);
    v[x_id][y_id].push_back(i + 1);
  }
  flag = false;
  for (int i = 0; i < sz; i++) {
    int st, step, ed;
    if (i & 1) {
      st = sz - 1, step = -1, ed = -1;
    } else {
      st = 0, step = 1, ed = sz;
    }
    for (int j = st; j != ed; j += step) {
      if (!v[i][j].size()) continue;
      sort(v[i][j].begin(), v[i][j].end(), cmp);
      for (int k = 0; k < v[i][j].size(); k++) {
        if (flag) printf(" ");
        printf("%d", v[i][j][k]);
        flag = true;
      }
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 max_n = 1111111;
const int edge = 500;
const int sz = 2000;
int n;
int x[max_n], y[max_n];
vector<int> v[sz][sz];
bool flag;
int cmp(const int a, const int b) {
  if (x[a] != x[b]) return x[a] < x[b];
  return y[a] < y[b];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    int x_id = min(x[i] / edge, sz - 1), y_id = min(y[i] / edge, sz - 1);
    v[x_id][y_id].push_back(i + 1);
  }
  flag = false;
  for (int i = 0; i < sz; i++) {
    int st, step, ed;
    if (i & 1) {
      st = sz - 1, step = -1, ed = -1;
    } else {
      st = 0, step = 1, ed = sz;
    }
    for (int j = st; j != ed; j += step) {
      if (!v[i][j].size()) continue;
      sort(v[i][j].begin(), v[i][j].end(), cmp);
      for (int k = 0; k < v[i][j].size(); k++) {
        if (flag) printf(" ");
        printf("%d", v[i][j][k]);
        flag = true;
      }
    }
  }
  printf("\n");
  return 0;
}

```