#include <bits/stdc++.h>
using namespace std;
void desperate_optimization(int precision) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
const int BLOCKS = 1e3;
const int N = 1e6;
pair<pair<int, int>, int> vec[N + 5];
vector<pair<pair<int, int>, int> > tmp;
bool cmp1(pair<pair<int, int>, int> &A, pair<pair<int, int>, int> &B) {
  return A.first.second < B.first.second;
}
bool cmp2(pair<pair<int, int>, int> &A, pair<pair<int, int>, int> &B) {
  return A.first.second > B.first.second;
}
void hora(int tc) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    vec[i] = {{x, y}, i + 1};
  }
  sort(vec, vec + n);
  int j = 0;
  for (int i = 0; i < n; i += BLOCKS) {
    tmp.clear();
    for (int j = i; j < min(n, i + BLOCKS); j++) tmp.push_back(vec[j]);
    if (i / BLOCKS % 2)
      sort(tmp.begin(), tmp.end(), cmp2);
    else
      sort(tmp.begin(), tmp.end(), cmp1);
    for (auto x : tmp) printf("%d ", x.second);
  }
  printf("\n");
}
int main() {
  desperate_optimization(10);
  int ntc = 1;
  for (int tc = 1; tc <= ntc; tc++) hora(tc);
  return 0;
}

### Prompt

Generate a cpp solution to 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;
void desperate_optimization(int precision) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
const int BLOCKS = 1e3;
const int N = 1e6;
pair<pair<int, int>, int> vec[N + 5];
vector<pair<pair<int, int>, int> > tmp;
bool cmp1(pair<pair<int, int>, int> &A, pair<pair<int, int>, int> &B) {
  return A.first.second < B.first.second;
}
bool cmp2(pair<pair<int, int>, int> &A, pair<pair<int, int>, int> &B) {
  return A.first.second > B.first.second;
}
void hora(int tc) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    vec[i] = {{x, y}, i + 1};
  }
  sort(vec, vec + n);
  int j = 0;
  for (int i = 0; i < n; i += BLOCKS) {
    tmp.clear();
    for (int j = i; j < min(n, i + BLOCKS); j++) tmp.push_back(vec[j]);
    if (i / BLOCKS % 2)
      sort(tmp.begin(), tmp.end(), cmp2);
    else
      sort(tmp.begin(), tmp.end(), cmp1);
    for (auto x : tmp) printf("%d ", x.second);
  }
  printf("\n");
}
int main() {
  desperate_optimization(10);
  int ntc = 1;
  for (int tc = 1; tc <= ntc; tc++) hora(tc);
  return 0;
}

```