#include <bits/stdc++.h>
using namespace std;
struct cp {
  int x, y;
} a[1000100];
vector<int> r[1010][1010];
bool bf[1010][1010];
bool cmp(int i, int j) {
  if (a[i].x == a[j].x) return a[i].y < a[j].y;
  return (a[i].x < a[j].x);
}
bool cmp2(int i, int j) {
  if (a[i].x == a[j].x) return a[i].y < a[j].y;
  return (a[i].x > a[j].x);
}
int n, i, j;
int dx[4] = {0, +1, 0, -1};
int dy[4] = {+1, 0, -1, 0};
int dir;
bool ok(int i, int j) {
  if (i < 0 || j < 0 || i > 1000 || j > 1000) return false;
  if (bf[i][j] == true) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    r[a[i].x / 1000][a[i].y / 1000].push_back(i);
  }
  for (i = 0; i <= 1000; i++) {
    if (i & 1)
      for (j = 1000; j >= 0; j--) {
        sort(r[i][j].begin(), r[i][j].end(), cmp2);
      }
    else
      for (j = 0; j <= 1000; j++) {
        sort(r[i][j].begin(), r[i][j].end(), cmp);
      }
  }
  i = 0;
  j = 0;
  dir = 0;
  while (true) {
    bf[i][j] = true;
    for (vector<int>::iterator it = r[i][j].begin(); it != r[i][j].end(); it++)
      printf("%d ", *it);
    if (ok(i + dx[dir], j + dy[dir]) == false) dir = (dir + 1) % 4;
    if (ok(i + dx[dir], j + dy[dir]) == false) break;
    i += dx[dir];
    j += dy[dir];
  }
}

### 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;
struct cp {
  int x, y;
} a[1000100];
vector<int> r[1010][1010];
bool bf[1010][1010];
bool cmp(int i, int j) {
  if (a[i].x == a[j].x) return a[i].y < a[j].y;
  return (a[i].x < a[j].x);
}
bool cmp2(int i, int j) {
  if (a[i].x == a[j].x) return a[i].y < a[j].y;
  return (a[i].x > a[j].x);
}
int n, i, j;
int dx[4] = {0, +1, 0, -1};
int dy[4] = {+1, 0, -1, 0};
int dir;
bool ok(int i, int j) {
  if (i < 0 || j < 0 || i > 1000 || j > 1000) return false;
  if (bf[i][j] == true) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    r[a[i].x / 1000][a[i].y / 1000].push_back(i);
  }
  for (i = 0; i <= 1000; i++) {
    if (i & 1)
      for (j = 1000; j >= 0; j--) {
        sort(r[i][j].begin(), r[i][j].end(), cmp2);
      }
    else
      for (j = 0; j <= 1000; j++) {
        sort(r[i][j].begin(), r[i][j].end(), cmp);
      }
  }
  i = 0;
  j = 0;
  dir = 0;
  while (true) {
    bf[i][j] = true;
    for (vector<int>::iterator it = r[i][j].begin(); it != r[i][j].end(); it++)
      printf("%d ", *it);
    if (ok(i + dx[dir], j + dy[dir]) == false) dir = (dir + 1) % 4;
    if (ok(i + dx[dir], j + dy[dir]) == false) break;
    i += dx[dir];
    j += dy[dir];
  }
}

```