#include <bits/stdc++.h>
using namespace std;
struct pt {
  int x, y;
  int i;
};
const int magic = 2500;
bool cmpX(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool cmpY(pt a, pt b) { return a.y < b.y || (a.y == b.y && a.x < b.x); }
bool cmpVP(pt a, pt b) {
  return (
      (a.x * 1LL * b.y - a.y * 1LL * b.x) > 0 ||
      (a.x * 1LL * b.y - a.y * 1LL * b.x == 0 &&
       a.x * 1LL * a.x + a.y * 1LL * a.y < b.x * 1LL * b.x + b.y * 1LL * b.y));
}
bool cmpVP2(pt a, pt b) {
  return (
      (a.x * 1LL * b.y - a.y * 1LL * b.x) < 0 ||
      (a.x * 1LL * b.y - a.y * 1LL * b.x == 0 &&
       a.x * 1LL * a.x + a.y * 1LL * a.y < b.x * 1LL * b.x + b.y * 1LL * b.y));
}
bool cmp1(pt a, pt b) {
  if (a.y / magic == b.y / magic) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
  } else {
    return (a.y / magic) < (b.y / magic);
  }
}
bool cmp2(pt a, pt b) {
  if (a.x / magic == b.x / magic) {
    return a.y < b.y || (a.y == b.y && a.x < b.x);
  } else {
    return (a.x / magic) < (b.x / magic);
  }
}
pt base, p[1000007];
int n;
long long res = 1e+18;
int ans[1000007];
void check() {
  long long dist = 0;
  for (int i = 1; i < n; i++)
    dist += ((p[i - 1].x - p[i].x) > 0 ? (p[i - 1].x - p[i].x)
                                       : -(p[i - 1].x - p[i].x)) +
            ((p[i - 1].y - p[i].y) > 0 ? (p[i - 1].y - p[i].y)
                                       : -(p[i - 1].y - p[i].y));
  if (dist < res) {
    for (int i = 0; i < n; i++) ans[i] = p[i].i + 1;
    res = dist;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].i = i;
  }
  sort(p, p + n, &cmpVP);
  check();
  sort(p, p + n, &cmpVP2);
  check();
  sort(p, p + n, &cmp1);
  check();
  sort(p, p + n, &cmp2);
  check();
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 pt {
  int x, y;
  int i;
};
const int magic = 2500;
bool cmpX(pt a, pt b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool cmpY(pt a, pt b) { return a.y < b.y || (a.y == b.y && a.x < b.x); }
bool cmpVP(pt a, pt b) {
  return (
      (a.x * 1LL * b.y - a.y * 1LL * b.x) > 0 ||
      (a.x * 1LL * b.y - a.y * 1LL * b.x == 0 &&
       a.x * 1LL * a.x + a.y * 1LL * a.y < b.x * 1LL * b.x + b.y * 1LL * b.y));
}
bool cmpVP2(pt a, pt b) {
  return (
      (a.x * 1LL * b.y - a.y * 1LL * b.x) < 0 ||
      (a.x * 1LL * b.y - a.y * 1LL * b.x == 0 &&
       a.x * 1LL * a.x + a.y * 1LL * a.y < b.x * 1LL * b.x + b.y * 1LL * b.y));
}
bool cmp1(pt a, pt b) {
  if (a.y / magic == b.y / magic) {
    return a.x < b.x || (a.x == b.x && a.y < b.y);
  } else {
    return (a.y / magic) < (b.y / magic);
  }
}
bool cmp2(pt a, pt b) {
  if (a.x / magic == b.x / magic) {
    return a.y < b.y || (a.y == b.y && a.x < b.x);
  } else {
    return (a.x / magic) < (b.x / magic);
  }
}
pt base, p[1000007];
int n;
long long res = 1e+18;
int ans[1000007];
void check() {
  long long dist = 0;
  for (int i = 1; i < n; i++)
    dist += ((p[i - 1].x - p[i].x) > 0 ? (p[i - 1].x - p[i].x)
                                       : -(p[i - 1].x - p[i].x)) +
            ((p[i - 1].y - p[i].y) > 0 ? (p[i - 1].y - p[i].y)
                                       : -(p[i - 1].y - p[i].y));
  if (dist < res) {
    for (int i = 0; i < n; i++) ans[i] = p[i].i + 1;
    res = dist;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].i = i;
  }
  sort(p, p + n, &cmpVP);
  check();
  sort(p, p + n, &cmpVP2);
  check();
  sort(p, p + n, &cmp1);
  check();
  sort(p, p + n, &cmp2);
  check();
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  return 0;
}

```