#include <bits/stdc++.h>
template <class T>
inline bool Chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
template <class T>
inline bool Chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <class T>
inline T read(T &x) {
  static int f;
  static char c;
  for (f = 1; !isdigit(c = getchar());)
    if (c == '-') f = -f;
  for (x = 0; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return x *= f;
}
template <class T>
inline void write(T x, char P = 0) {
  static char s[20];
  static int top = 0;
  if (x < 0) x = -x, putchar('-');
  do s[++top] = x % 10 + 48;
  while (x /= 10);
  do putchar(s[top]);
  while (--top);
  if (P) putchar(P);
}
inline void open() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
const int MO = 1e9 + 7;
const int oo = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
const int MAXM = 1e6 + 5;
using namespace std;
int N;
struct Point {
  int x, y, id;
  bool operator<(const Point &A) const { return x - A.x ? x < A.x : y < A.y; }
} P[MAXM];
int main() {
  read(N);
  for (int i = 1, x, y; i <= N; ++i) {
    read(x), read(y);
    P[i] = (Point){x / 1400, y, i};
  }
  sort(P + 1, P + N + 1);
  for (int i = 1; i <= N; ++i) {
    write(P[i].id, ' ');
  }
  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>
template <class T>
inline bool Chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
template <class T>
inline bool Chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <class T>
inline T read(T &x) {
  static int f;
  static char c;
  for (f = 1; !isdigit(c = getchar());)
    if (c == '-') f = -f;
  for (x = 0; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return x *= f;
}
template <class T>
inline void write(T x, char P = 0) {
  static char s[20];
  static int top = 0;
  if (x < 0) x = -x, putchar('-');
  do s[++top] = x % 10 + 48;
  while (x /= 10);
  do putchar(s[top]);
  while (--top);
  if (P) putchar(P);
}
inline void open() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
const int MO = 1e9 + 7;
const int oo = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
const int MAXM = 1e6 + 5;
using namespace std;
int N;
struct Point {
  int x, y, id;
  bool operator<(const Point &A) const { return x - A.x ? x < A.x : y < A.y; }
} P[MAXM];
int main() {
  read(N);
  for (int i = 1, x, y; i <= N; ++i) {
    read(x), read(y);
    P[i] = (Point){x / 1400, y, i};
  }
  sort(P + 1, P + N + 1);
  for (int i = 1; i <= N; ++i) {
    write(P[i].id, ' ');
  }
  return 0;
}

```