#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
struct point {
  int x, y, id;
  bool operator<(const point oth) const {
    if (x < oth.x) return true;
    if (x == oth.x) {
      if (y < oth.y) return true;
    }
    return false;
  }
};
vector<vector<point> > gr;
int n;
int ans[10000], en;
void show(int q) {
  if (q) {
    for (int i = 0; i < en; ++i) {
      printf("%d ", ans[i]);
    }
  } else {
    for (int i = en - 1; i >= 0; --i) {
      printf("%d ", ans[i]);
    }
  }
  en = 0;
}
void solve() {
  gr.resize(1001);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    point temp;
    scanf("%d %d", &temp.x, &temp.y);
    temp.id = i + 1;
    gr[temp.y / 1000].push_back(temp);
  }
  for (int i = 0; i <= 1000; ++i) {
    if (!gr[i].size()) continue;
    sort(gr[i].begin(), gr[i].end());
    int q = 0, pr = -1;
    if (i & 1) {
      q = 1;
      for (int j = gr[i].size() - 1; j >= 0; --j) {
        if (gr[i][j].x != pr) {
          show(q);
          pr = gr[i][j].x;
          q ^= 1;
        }
        ans[en++] = gr[i][j].id;
      }
    } else {
      for (int j = 0; j < gr[i].size(); ++j) {
        if (gr[i][j].x != pr) {
          show(q);
          pr = gr[i][j].x;
          q ^= 1;
        }
        ans[en++] = gr[i][j].id;
      }
    }
    show(q);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

### Prompt

Your challenge is to write a CPP solution 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 N = 1000010;
struct point {
  int x, y, id;
  bool operator<(const point oth) const {
    if (x < oth.x) return true;
    if (x == oth.x) {
      if (y < oth.y) return true;
    }
    return false;
  }
};
vector<vector<point> > gr;
int n;
int ans[10000], en;
void show(int q) {
  if (q) {
    for (int i = 0; i < en; ++i) {
      printf("%d ", ans[i]);
    }
  } else {
    for (int i = en - 1; i >= 0; --i) {
      printf("%d ", ans[i]);
    }
  }
  en = 0;
}
void solve() {
  gr.resize(1001);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    point temp;
    scanf("%d %d", &temp.x, &temp.y);
    temp.id = i + 1;
    gr[temp.y / 1000].push_back(temp);
  }
  for (int i = 0; i <= 1000; ++i) {
    if (!gr[i].size()) continue;
    sort(gr[i].begin(), gr[i].end());
    int q = 0, pr = -1;
    if (i & 1) {
      q = 1;
      for (int j = gr[i].size() - 1; j >= 0; --j) {
        if (gr[i][j].x != pr) {
          show(q);
          pr = gr[i][j].x;
          q ^= 1;
        }
        ans[en++] = gr[i][j].id;
      }
    } else {
      for (int j = 0; j < gr[i].size(); ++j) {
        if (gr[i][j].x != pr) {
          show(q);
          pr = gr[i][j].x;
          q ^= 1;
        }
        ans[en++] = gr[i][j].id;
      }
    }
    show(q);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

```