#include <bits/stdc++.h>
using namespace std;
struct lyrics {
  string line1;
  string line2;
};
struct myp {
  int f;
  int s;
};
struct edges {
  int v;
  bool check1;
};
struct mypair {
  string f;
  string s;
};
int dx[5] = {-1, 0, 0, 1, 0};
int dy[5] = {0, 1, -1, 0, 0};
int countvowel[100007];
bool isvowel[29];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
bool sorting(myp a, myp b) { return a.f < b.f; };
bool isok(int n, int m, int x, int y, char** a) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    return true;
    return false;
  }
  return false;
}
int gcd(int num, int denum) {
  if (num == 0)
    return denum;
  else
    return gcd(denum % num, num);
}
int get(long long x, int& finalzero) {
  int len = 0;
  while (x != 0) {
    len += 1;
    if (x % 2 == 0) {
      if (len <= 30) finalzero = len;
    }
    x /= 2;
  }
  return len;
}
int myabs(int a) {
  if (a >= 0)
    return a;
  else
    return -a;
}
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1" << endl;
    cout << "1 1" << endl;
    return 0;
  }
  int m = n / 2 + 1;
  cout << m << endl;
  int count = 0;
  for (int i = 0; i < m; i++) {
    cout << "1 " << i + 1 << endl;
    count += 1;
  }
  for (int i = 1; i < m; i++) {
    if (count == n) break;
    cout << i + 1 << " " << m << endl;
    count += 1;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct lyrics {
  string line1;
  string line2;
};
struct myp {
  int f;
  int s;
};
struct edges {
  int v;
  bool check1;
};
struct mypair {
  string f;
  string s;
};
int dx[5] = {-1, 0, 0, 1, 0};
int dy[5] = {0, 1, -1, 0, 0};
int countvowel[100007];
bool isvowel[29];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
bool sorting(myp a, myp b) { return a.f < b.f; };
bool isok(int n, int m, int x, int y, char** a) {
  if (x >= 0 && x < n && y >= 0 && y < m) {
    return true;
    return false;
  }
  return false;
}
int gcd(int num, int denum) {
  if (num == 0)
    return denum;
  else
    return gcd(denum % num, num);
}
int get(long long x, int& finalzero) {
  int len = 0;
  while (x != 0) {
    len += 1;
    if (x % 2 == 0) {
      if (len <= 30) finalzero = len;
    }
    x /= 2;
  }
  return len;
}
int myabs(int a) {
  if (a >= 0)
    return a;
  else
    return -a;
}
int main() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << "1" << endl;
    cout << "1 1" << endl;
    return 0;
  }
  int m = n / 2 + 1;
  cout << m << endl;
  int count = 0;
  for (int i = 0; i < m; i++) {
    cout << "1 " << i + 1 << endl;
    count += 1;
  }
  for (int i = 1; i < m; i++) {
    if (count == n) break;
    cout << i + 1 << " " << m << endl;
    count += 1;
  }
}

```