#include <bits/stdc++.h>
using namespace std;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
void setIO(string name) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin.exceptions(istream::failbit);
}
const int inf = 0x3f3f3f3f, mod = 1e9 + 7;
template <class T>
void print(T a, string sep = " ", string end = "\n") {
  for (auto i : a) {
    cout << i << sep;
  }
  cout << end;
}
int main() {
  setIO("1");
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
  }
  sort(a.begin(), a.end());
  int s = 1000;
  vector<int> sol;
  int en = s;
  for (int i = 0; i < n; en += s) {
    vector<pair<pair<int, int>, int>> cur;
    while (i < n && a[i].first.first <= en) {
      cur.push_back(a[i]);
      ++i;
    }
    sort(cur.begin(), cur.end(),
         [](pair<pair<int, int>, int> left, pair<pair<int, int>, int> right) {
           return left.first.second == right.first.second
                      ? left.first.first < right.first.first
                      : left.first.second < right.first.second;
         });
    for (auto j : cur) {
      sol.push_back(j.second);
    }
  }
  print(sol);
  return 0;
}

### Prompt

Your task is to create 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;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
void setIO(string name) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin.exceptions(istream::failbit);
}
const int inf = 0x3f3f3f3f, mod = 1e9 + 7;
template <class T>
void print(T a, string sep = " ", string end = "\n") {
  for (auto i : a) {
    cout << i << sep;
  }
  cout << end;
}
int main() {
  setIO("1");
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
  }
  sort(a.begin(), a.end());
  int s = 1000;
  vector<int> sol;
  int en = s;
  for (int i = 0; i < n; en += s) {
    vector<pair<pair<int, int>, int>> cur;
    while (i < n && a[i].first.first <= en) {
      cur.push_back(a[i]);
      ++i;
    }
    sort(cur.begin(), cur.end(),
         [](pair<pair<int, int>, int> left, pair<pair<int, int>, int> right) {
           return left.first.second == right.first.second
                      ? left.first.first < right.first.first
                      : left.first.second < right.first.second;
         });
    for (auto j : cur) {
      sol.push_back(j.second);
    }
  }
  print(sol);
  return 0;
}

```