#include <bits/stdc++.h>
using namespace std;
using K = long double;
using PII = pair<int, int>;
using PKK = pair<K, K>;
const int N = 1e6;
PII p[N];
map<PII, int> M;
bool xgty(PII a) { return a.first > a.second; }
K dst(PII xy, PKK a) {
  return abs(xy.first - a.first) + abs(xy.second - a.second);
}
void order(PII* begin, PII* end, PKK a, PKK b, PKK c) {
  if (end - begin <= 1) return;
  PII* ptr =
      partition(begin, end, [a, c](PII xy) { return dst(xy, a) < dst(xy, c); });
  PKK d = make_pair(0.5L * (a.first + c.first), 0.5L * (a.second + c.second));
  order(begin, ptr, a, d, b);
  order(ptr, end, b, d, c);
}
int main() {
  int n;
  assert(scanf("%d", &n));
  for (int i = 0; i < (n); i++) {
    assert(scanf("%d%d", &p[i].first, &p[i].second) == 2);
    M[p[i]] = i;
  }
  int k = partition(p, p + n, xgty) - p;
  order(p, p + k, make_pair(-1e6L, -1e6L), make_pair(1e6L, -1e6L),
        make_pair(1e6L, 1e6L));
  order(p + k, p + n, make_pair(-1e6L, -1e6L), make_pair(-1e6L, 1e6L),
        make_pair(1e6L, 1e6L));
  for (int i = 0; i < (n); i++) printf("%d ", M[p[i]] + 1);
  printf("\n");
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using K = long double;
using PII = pair<int, int>;
using PKK = pair<K, K>;
const int N = 1e6;
PII p[N];
map<PII, int> M;
bool xgty(PII a) { return a.first > a.second; }
K dst(PII xy, PKK a) {
  return abs(xy.first - a.first) + abs(xy.second - a.second);
}
void order(PII* begin, PII* end, PKK a, PKK b, PKK c) {
  if (end - begin <= 1) return;
  PII* ptr =
      partition(begin, end, [a, c](PII xy) { return dst(xy, a) < dst(xy, c); });
  PKK d = make_pair(0.5L * (a.first + c.first), 0.5L * (a.second + c.second));
  order(begin, ptr, a, d, b);
  order(ptr, end, b, d, c);
}
int main() {
  int n;
  assert(scanf("%d", &n));
  for (int i = 0; i < (n); i++) {
    assert(scanf("%d%d", &p[i].first, &p[i].second) == 2);
    M[p[i]] = i;
  }
  int k = partition(p, p + n, xgty) - p;
  order(p, p + k, make_pair(-1e6L, -1e6L), make_pair(1e6L, -1e6L),
        make_pair(1e6L, 1e6L));
  order(p + k, p + n, make_pair(-1e6L, -1e6L), make_pair(-1e6L, 1e6L),
        make_pair(1e6L, 1e6L));
  for (int i = 0; i < (n); i++) printf("%d ", M[p[i]] + 1);
  printf("\n");
}

```