#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int mod = 1e9 + 7;
const long long int LONGINF = 4e18;
const double eps = 1e-9;
const double PI = 3.1415926535897932384626433832795;
struct p {
  int a, b, n;
};
bool srt(p a, p b) {
  if (a.a / 1000 != b.a / 1000) return a.a < b.a;
  return a.b < b.b;
}
bool srt1(p a, p b) { return a.a < b.a; }
void solve(int n) {
  vector<p> v;
  p a;
  for (int i = 0; i < (n); i++) {
    scanf("%d%d", &a.a, &a.b);
    a.n = i + 1;
    v.push_back(a);
  }
  sort((v).begin(), (v).end(), srt);
  bool rv = false;
  for (int i = 0; i < (n); i++) {
    int l = i, r = i;
    while (r < n && v[l].a / 1000 == v[r].a / 1000) {
      r++;
      i++;
    }
    i--;
    if (rv) {
      reverse(v.begin() + l, v.begin() + r);
      rv = false;
    } else
      rv = true;
  }
  rv = false;
  for (int i = 0; i < (n); i++) {
    int l = i, r = i;
    while (r < n && v[l].a / 1000 == v[r].a / 1000 && v[l].b == v[r].b) {
      r++;
      i++;
    }
    i--;
    sort(v.begin() + l, v.begin() + r, srt1);
    if (rv) {
      reverse(v.begin() + l, v.begin() + r);
      rv = false;
    } else
      rv = true;
  }
  for (int i = 0; i < (n); i++) {
    printf("%d ", v[i].n);
  }
}
int main() {
#pragma comment(linker, "/STACK:268435456")
  int n;
  string s;
  while (scanf("%d", &n) == 1) solve(n);
  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;
const int INF = 1e9;
const int mod = 1e9 + 7;
const long long int LONGINF = 4e18;
const double eps = 1e-9;
const double PI = 3.1415926535897932384626433832795;
struct p {
  int a, b, n;
};
bool srt(p a, p b) {
  if (a.a / 1000 != b.a / 1000) return a.a < b.a;
  return a.b < b.b;
}
bool srt1(p a, p b) { return a.a < b.a; }
void solve(int n) {
  vector<p> v;
  p a;
  for (int i = 0; i < (n); i++) {
    scanf("%d%d", &a.a, &a.b);
    a.n = i + 1;
    v.push_back(a);
  }
  sort((v).begin(), (v).end(), srt);
  bool rv = false;
  for (int i = 0; i < (n); i++) {
    int l = i, r = i;
    while (r < n && v[l].a / 1000 == v[r].a / 1000) {
      r++;
      i++;
    }
    i--;
    if (rv) {
      reverse(v.begin() + l, v.begin() + r);
      rv = false;
    } else
      rv = true;
  }
  rv = false;
  for (int i = 0; i < (n); i++) {
    int l = i, r = i;
    while (r < n && v[l].a / 1000 == v[r].a / 1000 && v[l].b == v[r].b) {
      r++;
      i++;
    }
    i--;
    sort(v.begin() + l, v.begin() + r, srt1);
    if (rv) {
      reverse(v.begin() + l, v.begin() + r);
      rv = false;
    } else
      rv = true;
  }
  for (int i = 0; i < (n); i++) {
    printf("%d ", v[i].n);
  }
}
int main() {
#pragma comment(linker, "/STACK:268435456")
  int n;
  string s;
  while (scanf("%d", &n) == 1) solve(n);
  return 0;
}

```