#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int m;
pair<pair<int, int>, int> q[N];
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
inline bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first == b.first.first) {
    if (a.first.first & 1)
      return a.first.second < b.first.second;
    else
      return a.first.second > b.first.second;
  } else
    return a.first.first < b.first.first;
}
int main() {
  scanf("%d", &m);
  int len = 1000;
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read();
    q[i] = make_pair(make_pair(l / len, r), i);
  }
  sort(&q[1], &q[m + 1], cmp);
  for (int i = 1; i <= m; i++) {
    write(q[i].second);
    putchar(' ');
  }
}

### Prompt

Please provide a Cpp coded solution 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 N = 1000005;
int m;
pair<pair<int, int>, int> q[N];
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(int a) {
  if (a >= 10) write(a / 10);
  putchar('0' + a % 10);
}
inline void writeln(int a) {
  if (a < 0) {
    a = -a;
    putchar('-');
  }
  write(a);
  puts("");
}
inline bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first == b.first.first) {
    if (a.first.first & 1)
      return a.first.second < b.first.second;
    else
      return a.first.second > b.first.second;
  } else
    return a.first.first < b.first.first;
}
int main() {
  scanf("%d", &m);
  int len = 1000;
  for (int i = 1; i <= m; i++) {
    int l = read(), r = read();
    q[i] = make_pair(make_pair(l / len, r), i);
  }
  sort(&q[1], &q[m + 1], cmp);
  for (int i = 1; i <= m; i++) {
    write(q[i].second);
    putchar(' ');
  }
}

```