#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
template <typename T>
bool chkmin(T &a, T b) {
  return b < a ? a = b, true : false;
}
template <typename T>
bool chkmax(T &a, T b) {
  return b > a ? a = b, true : false;
}
template <class T>
void read(T &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 * 10 + c - '0';
  x *= flag;
}
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  static char s[20];
  int top = 0;
  while (x) s[++top] = x % 10 + '0', x /= 10;
  if (top == 0) putchar('0');
  while (top) putchar(s[top--]);
}
const int MAXN = 1e6 + 100;
const int S = 1000;
struct Q {
  int x, y, id;
  friend bool operator<(const Q A, const Q B) {
    if (A.x / S != B.x / S) return A.x < B.x;
    return ((A.x / S) & 1) ^ (A.y < B.y);
  }
} a[MAXN];
int N;
int main() {
  read(N);
  for (int i = 1; i <= N; ++i) {
    read(a[i].x), read(a[i].y), a[i].id = i;
  }
  sort(a + 1, a + N + 1);
  for (int i = 1; i <= N; ++i) write(a[i].id), putchar(' ');
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int oo = 0x3f3f3f3f;
template <typename T>
bool chkmin(T &a, T b) {
  return b < a ? a = b, true : false;
}
template <typename T>
bool chkmax(T &a, T b) {
  return b > a ? a = b, true : false;
}
template <class T>
void read(T &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 * 10 + c - '0';
  x *= flag;
}
template <class T>
void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  static char s[20];
  int top = 0;
  while (x) s[++top] = x % 10 + '0', x /= 10;
  if (top == 0) putchar('0');
  while (top) putchar(s[top--]);
}
const int MAXN = 1e6 + 100;
const int S = 1000;
struct Q {
  int x, y, id;
  friend bool operator<(const Q A, const Q B) {
    if (A.x / S != B.x / S) return A.x < B.x;
    return ((A.x / S) & 1) ^ (A.y < B.y);
  }
} a[MAXN];
int N;
int main() {
  read(N);
  for (int i = 1; i <= N; ++i) {
    read(a[i].x), read(a[i].y), a[i].id = i;
  }
  sort(a + 1, a + N + 1);
  for (int i = 1; i <= N; ++i) write(a[i].id), putchar(' ');
  return 0;
}

```