#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
vector<long long> vv;
set<long long> st;
long long x, y, z;
string s, s1, s2, s3, s4, s5;
void memset_ll(long long a[], long long v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
void memset_s(char a[], char v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
int compare(const void* a, const void* b) {
  return (*(long long*)a - *(long long*)b);
}
void q_sort(long long a[], long long n) {
  qsort(a, n, sizeof(long long), compare);
}
static int compare1(const void* a, const void* b) {
  if (*(double*)a > *(double*)b)
    return 1;
  else if (*(double*)a < *(double*)b)
    return -1;
  else
    return 0;
}
void q_sort_d(double a[], long long n) {
  qsort(a, n, sizeof(double), compare1);
}
string trans_upper(string ss) {
  transform(ss.begin(), ss.end(), ss.begin(), ::toupper);
  return ss;
}
long long power(long long a, long long b) {
  long long pow = 1;
  while (b) {
    if (b & 1) {
      pow = pow * a;
      --b;
    }
    a = a * a;
    b = b / 2;
  }
  return pow;
}
long long bsearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return bsearch(arr, l, mid - 1, x);
    return bsearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long fact(long long n) {
  long long f = 1, i;
  for (i = 1; i <= n; i++) f = f * i;
  return f;
}
long long is_prime(long long n) {
  long long i, j, k = 1;
  if (n % 2 == 0) {
    k = 0;
    return k;
  }
  j = sqrt(n);
  for (i = 3; i <= j; i += 2) {
    if (n % i == 0) {
      k = 0;
      break;
    }
  }
  return k;
}
long long vow(char c) {
  long long k = 1;
  long long l = 0;
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
    return k;
  else
    return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, l, p, q, n, m, o = 0, r, i, j, h, k, e = 0, d = 0, f = 0, g = 0;
  long long temp = 0, nl, np;
  double p1, p2 = 0.0, p3 = 0.0, p4, p5, p6, pp = 0.0, ppp;
  p1 = 3.1415926536;
  long long x1, x2, y1, y2, uu = 0, dd = 0, rr = 0, lll = 0;
  cin >> n;
  if (n == 1) {
    cout << 1 << endl;
    cout << 1 << " " << 1;
    return 0;
  }
  if (n % 2 == 0) {
    p = 1;
    q = 1;
    r = n / 2;
    h = r;
    if (n % 2 != 0) h++;
    t = p + h;
    cout << t << endl;
    for (i = 1; i <= h; i++) {
      cout << 1 << " " << i << endl;
    }
    for (i = t - r + 1; i <= t; i++) {
      cout << t << " " << i << endl;
    }
    return 0;
  } else {
    p = 1;
    q = 1;
    n--;
    r = n / 2;
    h = r + 1;
    t = p + r;
    cout << t << endl;
    for (i = 1; i <= h; i++) {
      cout << 1 << " " << i << endl;
    }
    for (i = t - r + 1; i <= t; i++) {
      cout << t << " " << i << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
vector<long long> v;
vector<long long> vv;
set<long long> st;
long long x, y, z;
string s, s1, s2, s3, s4, s5;
void memset_ll(long long a[], long long v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
void memset_s(char a[], char v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
int compare(const void* a, const void* b) {
  return (*(long long*)a - *(long long*)b);
}
void q_sort(long long a[], long long n) {
  qsort(a, n, sizeof(long long), compare);
}
static int compare1(const void* a, const void* b) {
  if (*(double*)a > *(double*)b)
    return 1;
  else if (*(double*)a < *(double*)b)
    return -1;
  else
    return 0;
}
void q_sort_d(double a[], long long n) {
  qsort(a, n, sizeof(double), compare1);
}
string trans_upper(string ss) {
  transform(ss.begin(), ss.end(), ss.begin(), ::toupper);
  return ss;
}
long long power(long long a, long long b) {
  long long pow = 1;
  while (b) {
    if (b & 1) {
      pow = pow * a;
      --b;
    }
    a = a * a;
    b = b / 2;
  }
  return pow;
}
long long bsearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return bsearch(arr, l, mid - 1, x);
    return bsearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long fact(long long n) {
  long long f = 1, i;
  for (i = 1; i <= n; i++) f = f * i;
  return f;
}
long long is_prime(long long n) {
  long long i, j, k = 1;
  if (n % 2 == 0) {
    k = 0;
    return k;
  }
  j = sqrt(n);
  for (i = 3; i <= j; i += 2) {
    if (n % i == 0) {
      k = 0;
      break;
    }
  }
  return k;
}
long long vow(char c) {
  long long k = 1;
  long long l = 0;
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
    return k;
  else
    return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, l, p, q, n, m, o = 0, r, i, j, h, k, e = 0, d = 0, f = 0, g = 0;
  long long temp = 0, nl, np;
  double p1, p2 = 0.0, p3 = 0.0, p4, p5, p6, pp = 0.0, ppp;
  p1 = 3.1415926536;
  long long x1, x2, y1, y2, uu = 0, dd = 0, rr = 0, lll = 0;
  cin >> n;
  if (n == 1) {
    cout << 1 << endl;
    cout << 1 << " " << 1;
    return 0;
  }
  if (n % 2 == 0) {
    p = 1;
    q = 1;
    r = n / 2;
    h = r;
    if (n % 2 != 0) h++;
    t = p + h;
    cout << t << endl;
    for (i = 1; i <= h; i++) {
      cout << 1 << " " << i << endl;
    }
    for (i = t - r + 1; i <= t; i++) {
      cout << t << " " << i << endl;
    }
    return 0;
  } else {
    p = 1;
    q = 1;
    n--;
    r = n / 2;
    h = r + 1;
    t = p + r;
    cout << t << endl;
    for (i = 1; i <= h; i++) {
      cout << 1 << " " << i << endl;
    }
    for (i = t - r + 1; i <= t; i++) {
      cout << t << " " << i << endl;
    }
  }
  return 0;
}

```