#include <bits/stdc++.h>
using namespace std;
struct node {
  int x, y, id;
};
node A[1000100];
int ans[1000100];
vector<int> P[1200];
int cmp1(node a, node b) { return a.x < b.x; }
int cmp2(int a, int b) { return A[a].y < A[b].y; }
const int Base = 1000;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d %d", &A[i].x, &A[i].y), A[i].id = i;
  sort(A + 1, A + n + 1, cmp1);
  for (int i = 1; i <= n; i++) {
    P[(A[i].x - 1) / Base + 1].push_back(i);
  }
  int k = 0;
  for (int i = 0; i <= 1018; i++) {
    sort(P[i].begin(), P[i].end(), cmp2);
    for (int j = 0; j < P[i].size(); j++) {
      ans[++k] = A[P[i][j]].id;
    }
  }
  for (int i = 1; i <= k; i++) printf("%d ", ans[i]);
  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;
struct node {
  int x, y, id;
};
node A[1000100];
int ans[1000100];
vector<int> P[1200];
int cmp1(node a, node b) { return a.x < b.x; }
int cmp2(int a, int b) { return A[a].y < A[b].y; }
const int Base = 1000;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d %d", &A[i].x, &A[i].y), A[i].id = i;
  sort(A + 1, A + n + 1, cmp1);
  for (int i = 1; i <= n; i++) {
    P[(A[i].x - 1) / Base + 1].push_back(i);
  }
  int k = 0;
  for (int i = 0; i <= 1018; i++) {
    sort(P[i].begin(), P[i].end(), cmp2);
    for (int j = 0; j < P[i].size(); j++) {
      ans[++k] = A[P[i][j]].id;
    }
  }
  for (int i = 1; i <= k; i++) printf("%d ", ans[i]);
  return 0;
}

```