#include <bits/stdc++.h>
using namespace std;
inline void file() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (0) {
    freopen(".in", "r", stdin);
    freopen(".out", "w", stdout);
  }
}
const clock_t MAXT = (100 * CLOCKS_PER_SEC) / 1000;
const int PX[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          PY[8] = {0, 1, 0, -1, -1, 1, 1, -1}, N = 1e6 + 10, MXCOR = 1e6,
          LOG = 18, INF = 1e9, MOD = 1e9 + 7;
const long long INFL = 1e18, MODL = 1e9 + 7;
const long double EPS = 1e-9;
inline int rnd(int l = 0, int r = INF) {
  unsigned ans = rand();
  ans = (ans << 8) ^ rand();
  ans = (ans << 8) ^ rand();
  ans = (ans << 8) ^ rand();
  ans %= r - l + 1;
  return int(ans + l);
}
int n, step, pos;
pair<pair<int, int>, int> a[N];
vector<pair<pair<int, int>, int>> v;
inline bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second != y.first.second) return x.first.second < y.first.second;
  return x < y;
}
int main() {
  file();
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
  }
  sort(a, a + n);
  step = 1000;
  pos = 0;
  for (int x = 0; x <= MXCOR; x += step) {
    v.clear();
    while (pos < n && a[pos].first.first <= x + step) v.push_back(a[pos++]);
    sort(v.begin(), v.end(), cmp);
    if ((x / step) % 2) reverse(v.begin(), v.end());
    for (auto i : v) cout << i.second << ' ';
  }
}

### Prompt

Generate 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;
inline void file() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (0) {
    freopen(".in", "r", stdin);
    freopen(".out", "w", stdout);
  }
}
const clock_t MAXT = (100 * CLOCKS_PER_SEC) / 1000;
const int PX[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          PY[8] = {0, 1, 0, -1, -1, 1, 1, -1}, N = 1e6 + 10, MXCOR = 1e6,
          LOG = 18, INF = 1e9, MOD = 1e9 + 7;
const long long INFL = 1e18, MODL = 1e9 + 7;
const long double EPS = 1e-9;
inline int rnd(int l = 0, int r = INF) {
  unsigned ans = rand();
  ans = (ans << 8) ^ rand();
  ans = (ans << 8) ^ rand();
  ans = (ans << 8) ^ rand();
  ans %= r - l + 1;
  return int(ans + l);
}
int n, step, pos;
pair<pair<int, int>, int> a[N];
vector<pair<pair<int, int>, int>> v;
inline bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  if (x.first.second != y.first.second) return x.first.second < y.first.second;
  return x < y;
}
int main() {
  file();
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
  }
  sort(a, a + n);
  step = 1000;
  pos = 0;
  for (int x = 0; x <= MXCOR; x += step) {
    v.clear();
    while (pos < n && a[pos].first.first <= x + step) v.push_back(a[pos++]);
    sort(v.begin(), v.end(), cmp);
    if ((x / step) % 2) reverse(v.begin(), v.end());
    for (auto i : v) cout << i.second << ' ';
  }
}

```