#include <bits/stdc++.h>
using namespace std;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
double Pi = acos(-1);
vector<pair<long long, long long> > v[901];
int main(void) {
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)(n); i++) {
    long long x, y;
    cin >> x >> y;
    double a = atan(1.0 * y / x) * 180.0 / Pi;
    long long j = a / 0.1;
    v[j].push_back(make_pair(x + y, i));
  }
  for (long long i = 0; i < (long long)(901); i++) {
    sort(v[i].begin(), v[i].end());
  }
  bool k = 1;
  for (long long i = 0; i < (long long)(901); i++) {
    if (v[i].size() == 0) continue;
    if (k) reverse(v[i].begin(), v[i].end());
    for (long long j = 0; j < (long long)(v[i].size()); j++) {
      cout << v[i][j].second + 1 << " ";
    }
    k = !k;
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
double Pi = acos(-1);
vector<pair<long long, long long> > v[901];
int main(void) {
  long long n;
  cin >> n;
  for (long long i = 0; i < (long long)(n); i++) {
    long long x, y;
    cin >> x >> y;
    double a = atan(1.0 * y / x) * 180.0 / Pi;
    long long j = a / 0.1;
    v[j].push_back(make_pair(x + y, i));
  }
  for (long long i = 0; i < (long long)(901); i++) {
    sort(v[i].begin(), v[i].end());
  }
  bool k = 1;
  for (long long i = 0; i < (long long)(901); i++) {
    if (v[i].size() == 0) continue;
    if (k) reverse(v[i].begin(), v[i].end());
    for (long long j = 0; j < (long long)(v[i].size()); j++) {
      cout << v[i][j].second + 1 << " ";
    }
    k = !k;
  }
  return 0;
}

```