#include <bits/stdc++.h>
using namespace std;
inline int rd() {
  register int res = 0, lab = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') lab = -lab;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + (ch & 15), ch = getchar();
  return res * lab;
}
inline void prt(int x, char tail = 0) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) prt(x / 10), x %= 10;
  putchar(x | 48);
  if (tail) putchar(tail);
}
struct node {
  int id, x, y;
  bool operator<(const node _cmp) const { return y < _cmp.y; }
  bool operator>(const node _cmp) const { return y > _cmp.y; }
};
bool cmp1(node a, node b) { return a > b; }
bool cmp2(node a, node b) { return a < b; }
int n;
node t;
vector<node> v[1005];
int main() {
  n = rd();
  for (int i = 1; i <= n; ++i) {
    t.id = i;
    t.x = rd();
    t.y = rd();
    v[t.x / 1000].push_back(t);
  }
  for (int i = 0; i <= 1000; ++i)
    sort(v[i].begin(), v[i].end(), (i & 1) ? (cmp1) : (cmp2));
  for (int i = 0; i <= 1000; ++i) {
    for (int j = 0; j < v[i].size(); ++j) prt(v[i][j].id, ' ');
  }
  fputs("\n", stdout);
  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;
inline int rd() {
  register int res = 0, lab = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') lab = -lab;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + (ch & 15), ch = getchar();
  return res * lab;
}
inline void prt(int x, char tail = 0) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) prt(x / 10), x %= 10;
  putchar(x | 48);
  if (tail) putchar(tail);
}
struct node {
  int id, x, y;
  bool operator<(const node _cmp) const { return y < _cmp.y; }
  bool operator>(const node _cmp) const { return y > _cmp.y; }
};
bool cmp1(node a, node b) { return a > b; }
bool cmp2(node a, node b) { return a < b; }
int n;
node t;
vector<node> v[1005];
int main() {
  n = rd();
  for (int i = 1; i <= n; ++i) {
    t.id = i;
    t.x = rd();
    t.y = rd();
    v[t.x / 1000].push_back(t);
  }
  for (int i = 0; i <= 1000; ++i)
    sort(v[i].begin(), v[i].end(), (i & 1) ? (cmp1) : (cmp2));
  for (int i = 0; i <= 1000; ++i) {
    for (int j = 0; j < v[i].size(); ++j) prt(v[i][j].id, ' ');
  }
  fputs("\n", stdout);
  return 0;
}

```