#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-10;
int dcmp(double x) {
  if (fabs(x) < eps)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 5;
struct node {
  int x, y, id;
  bool operator<(const node& u) const {
    return x < u.x || (x == u.x && y < u.y) ||
           (x == u.x && y == u.y && id < u.id);
  }
};
vector<node> G[1000 * 1000 + 2000];
int main() {
  int n, m, x, y;
  scanf("%d", &n);
  m = 1001;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    int xx = x / 1000, yy = y / 1000;
    G[yy * m + xx].push_back({x, y, i});
  }
  for (int i = 0; i <= 1000; i++) {
    if (i % 2 == 0) {
      for (int j = 0; j <= 1000; j++) {
        sort(G[i * m + j].begin(), G[i * m + j].end());
        for (node u : G[i * m + j]) {
          printf("%d ", u.id + 1);
        }
      }
    } else {
      for (int j = 1000; j >= 0; j--) {
        sort(G[i * m + j].begin(), G[i * m + j].end());
        for (int k = G[i * m + j].size() - 1; k >= 0; k--) {
          printf("%d ", G[i * m + j][k].id + 1);
        }
      }
    }
  }
  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 double PI = acos(-1.0);
const double eps = 1e-10;
int dcmp(double x) {
  if (fabs(x) < eps)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 5;
struct node {
  int x, y, id;
  bool operator<(const node& u) const {
    return x < u.x || (x == u.x && y < u.y) ||
           (x == u.x && y == u.y && id < u.id);
  }
};
vector<node> G[1000 * 1000 + 2000];
int main() {
  int n, m, x, y;
  scanf("%d", &n);
  m = 1001;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    int xx = x / 1000, yy = y / 1000;
    G[yy * m + xx].push_back({x, y, i});
  }
  for (int i = 0; i <= 1000; i++) {
    if (i % 2 == 0) {
      for (int j = 0; j <= 1000; j++) {
        sort(G[i * m + j].begin(), G[i * m + j].end());
        for (node u : G[i * m + j]) {
          printf("%d ", u.id + 1);
        }
      }
    } else {
      for (int j = 1000; j >= 0; j--) {
        sort(G[i * m + j].begin(), G[i * m + j].end());
        for (int k = G[i * m + j].size() - 1; k >= 0; k--) {
          printf("%d ", G[i * m + j][k].id + 1);
        }
      }
    }
  }
  return 0;
}

```