#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long gcd(long long a, long long b) { return (b) ? gcd(b, a % b) : a; }
long long modpow(long long n, long long k, long long mod) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    k >>= 1;
    n = (n * n) % mod;
  }
  return ans % mod;
}
long long ab(long long a) { return (a > 0) ? a : -a; }
long long lcm(long long a, long long b) {
  return (a * b) / gcd(max(a, b), min(a, b));
}
long long poww(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n);
    k >>= 1;
    n = (n * n);
  }
  return ans;
}
vector<pair<pair<int, int>, int> > V;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.first == y.first.first) {
    if ((x.first.first) & 1) {
      return x.first.second > y.first.second;
    } else {
      return x.first.second < y.first.second;
    }
  }
  return x.first.first < y.first.first;
}
int main() {
  int n, i, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &x, &y);
    V.push_back(make_pair(make_pair(x / (1000), y), i + 1));
  }
  sort(V.begin(), V.end(), cmp);
  for (i = 0; i < V.size(); i++) printf("%d ", V[i].second);
  return 0;
}

### 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;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long gcd(long long a, long long b) { return (b) ? gcd(b, a % b) : a; }
long long modpow(long long n, long long k, long long mod) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    k >>= 1;
    n = (n * n) % mod;
  }
  return ans % mod;
}
long long ab(long long a) { return (a > 0) ? a : -a; }
long long lcm(long long a, long long b) {
  return (a * b) / gcd(max(a, b), min(a, b));
}
long long poww(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n);
    k >>= 1;
    n = (n * n);
  }
  return ans;
}
vector<pair<pair<int, int>, int> > V;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.first == y.first.first) {
    if ((x.first.first) & 1) {
      return x.first.second > y.first.second;
    } else {
      return x.first.second < y.first.second;
    }
  }
  return x.first.first < y.first.first;
}
int main() {
  int n, i, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &x, &y);
    V.push_back(make_pair(make_pair(x / (1000), y), i + 1));
  }
  sort(V.begin(), V.end(), cmp);
  for (i = 0; i < V.size(); i++) printf("%d ", V[i].second);
  return 0;
}

```