#include <bits/stdc++.h>
const int N = 1000000;
int n = 0;
int p[N + 100];
struct node {
  long long y;
  long long x;
  long long k;
  int i;
};
node nodes[N + 100];
void get_ll(long long& r) {
  r = 0LL;
  char c = 0;
  do {
    c = getchar();
  } while (c == ' ' || c == '\n' || c == '\t');
  do {
    r *= 10LL;
    r += (c - '0');
    c = getchar();
  } while (c >= '0' && c <= '9');
}
bool comp(const node& a, const node& b) {
  if (a.k != b.k) return a.k < b.k;
  if (a.k % 2 == 0) return a.y < b.y;
  return a.y > b.y;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    get_ll(nodes[i].x);
    get_ll(nodes[i].y);
    nodes[i].k = nodes[i].x / 1000LL;
    nodes[i].i = i + 1;
  }
  std::sort(nodes, nodes + n, comp);
  for (int i = 0; i < n; i++) printf("%d ", nodes[i].i);
  printf("\n");
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const int N = 1000000;
int n = 0;
int p[N + 100];
struct node {
  long long y;
  long long x;
  long long k;
  int i;
};
node nodes[N + 100];
void get_ll(long long& r) {
  r = 0LL;
  char c = 0;
  do {
    c = getchar();
  } while (c == ' ' || c == '\n' || c == '\t');
  do {
    r *= 10LL;
    r += (c - '0');
    c = getchar();
  } while (c >= '0' && c <= '9');
}
bool comp(const node& a, const node& b) {
  if (a.k != b.k) return a.k < b.k;
  if (a.k % 2 == 0) return a.y < b.y;
  return a.y > b.y;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    get_ll(nodes[i].x);
    get_ll(nodes[i].y);
    nodes[i].k = nodes[i].x / 1000LL;
    nodes[i].i = i + 1;
  }
  std::sort(nodes, nodes + n, comp);
  for (int i = 0; i < n; i++) printf("%d ", nodes[i].i);
  printf("\n");
  return 0;
}

```