#include <bits/stdc++.h>
template <class Num>
void read(Num &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c - '0');
  x *= flag;
}
template <class Num>
void write(Num x) {
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  static char s[20];
  int sl = 0;
  while (x) s[sl++] = x % 10 + '0', x /= 10;
  while (sl) putchar(s[--sl]);
}
const int maxn = 1e6 + 20;
const int size = 1e3;
int n;
std::pair<std::pair<int, int>, int> rev[maxn];
bool cmp(const std::pair<std::pair<int, int>, int> &a,
         const std::pair<std::pair<int, int>, int> &b) {
  return a.first.second < b.first.second;
}
int main() {
  read(n);
  for (int i = 1, x, y; i <= n; i++) {
    read(x), read(y);
    rev[i] = std::make_pair(std::make_pair(y, x), i);
  }
  std::sort(rev + 1, rev + n + 1, cmp);
  for (int i = 1, pos = 1; i <= size; i++) {
    int st = pos;
    while (pos <= n && rev[pos].first.second <= i * size) pos++;
    if (i & 1)
      std::sort(rev + st, rev + pos);
    else
      std::sort(rev + st, rev + pos,
                std::greater<std::pair<std::pair<int, int>, int> >());
  }
  for (int i = 1; i <= n; i++) write(rev[i].second), putchar(' ');
  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>
template <class Num>
void read(Num &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c - '0');
  x *= flag;
}
template <class Num>
void write(Num x) {
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  static char s[20];
  int sl = 0;
  while (x) s[sl++] = x % 10 + '0', x /= 10;
  while (sl) putchar(s[--sl]);
}
const int maxn = 1e6 + 20;
const int size = 1e3;
int n;
std::pair<std::pair<int, int>, int> rev[maxn];
bool cmp(const std::pair<std::pair<int, int>, int> &a,
         const std::pair<std::pair<int, int>, int> &b) {
  return a.first.second < b.first.second;
}
int main() {
  read(n);
  for (int i = 1, x, y; i <= n; i++) {
    read(x), read(y);
    rev[i] = std::make_pair(std::make_pair(y, x), i);
  }
  std::sort(rev + 1, rev + n + 1, cmp);
  for (int i = 1, pos = 1; i <= size; i++) {
    int st = pos;
    while (pos <= n && rev[pos].first.second <= i * size) pos++;
    if (i & 1)
      std::sort(rev + st, rev + pos);
    else
      std::sort(rev + st, rev + pos,
                std::greater<std::pair<std::pair<int, int>, int> >());
  }
  for (int i = 1; i <= n; i++) write(rev[i].second), putchar(' ');
  return 0;
}

```