#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
struct Po {
  int x, y, b, id;
  Po() {}
  Po(int x, int y, int b, int id) : x(x), y(y), b(b), id(id) {}
  bool operator<(const Po& th) const {
    if (b != th.b) return b < th.b;
    if (y != th.y) return y < th.y;
    return x < th.x;
  }
} po[N];
int aa[N], bb[N];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int l = 1000000, r = 0;
    for (int i = 0; i < n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      aa[i] = a, bb[i] = b;
      l = min(l, a);
      r = max(r, a);
    }
    int t = (int)floor(sqrt((double)(r - l) + 1));
    for (int i = 0; i < n; i++) {
      int a, b;
      a = aa[i], b = bb[i];
      po[i] = Po(a, b, (a - l) / t, i);
    }
    sort(po, po + n);
    int i = 0;
    t = 0;
    int flag = 0;
    while (i < n) {
      int j = i;
      while (i < n - 1 && po[i].b == po[i + 1].b) i++;
      if (t == 0) {
        for (int k = j; k <= i; k++) {
          if (flag) printf(" ");
          printf("%d", po[k].id + 1);
          flag = 1;
        }
      } else {
        for (int k = i; k >= j; k--) {
          if (flag) printf(" ");
          printf("%d", po[k].id + 1);
          flag = 1;
        }
      }
      t ^= 1;
      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>
using namespace std;
const int N = 1e6 + 100;
struct Po {
  int x, y, b, id;
  Po() {}
  Po(int x, int y, int b, int id) : x(x), y(y), b(b), id(id) {}
  bool operator<(const Po& th) const {
    if (b != th.b) return b < th.b;
    if (y != th.y) return y < th.y;
    return x < th.x;
  }
} po[N];
int aa[N], bb[N];
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int l = 1000000, r = 0;
    for (int i = 0; i < n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      aa[i] = a, bb[i] = b;
      l = min(l, a);
      r = max(r, a);
    }
    int t = (int)floor(sqrt((double)(r - l) + 1));
    for (int i = 0; i < n; i++) {
      int a, b;
      a = aa[i], b = bb[i];
      po[i] = Po(a, b, (a - l) / t, i);
    }
    sort(po, po + n);
    int i = 0;
    t = 0;
    int flag = 0;
    while (i < n) {
      int j = i;
      while (i < n - 1 && po[i].b == po[i + 1].b) i++;
      if (t == 0) {
        for (int k = j; k <= i; k++) {
          if (flag) printf(" ");
          printf("%d", po[k].id + 1);
          flag = 1;
        }
      } else {
        for (int k = i; k >= j; k--) {
          if (flag) printf(" ");
          printf("%d", po[k].id + 1);
          flag = 1;
        }
      }
      t ^= 1;
      i++;
    }
    printf("\n");
  }
  return 0;
}

```