#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e5 + 7;
const long long MOD = 1e9 + 7;
long long modExp(long long base, long long power) {
  if (power == 0) {
    return 1;
  } else {
    long long cur = modExp(base, power / 2);
    cur = cur * cur;
    cur = cur % MOD;
    if (power % 2 == 1) cur = cur * base;
    cur = cur % MOD;
    return cur;
  }
}
long long inv(long long base) {
  int g = MOD, r = base, x = 0, y = 1;
  while (r != 0) {
    int q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return x < 0 ? x + MOD : x;
}
long long mul(long long A, long long B) { return (A * B) % MOD; }
long long add(long long A, long long B) { return (A + B) % MOD; }
long long dvd(long long A, long long B) { return mul(A, inv(B)); }
long long sub(long long A, long long B) { return (A - B + MOD) % MOD; }
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
void findPowerSet(vector<int> const &S, int n) {
  int N = pow(2, n);
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) cout << S[j] << " ";
    }
    cout << endl;
  }
}
long long findSubarraySum(vector<int> arr, int n) {
  vector<long long> prevSum;
  long long res = 0;
  int currN = 0;
  long long currsum = 0;
  for (int i = 0; i < n; i++) {
    currsum += arr[i];
    if (currsum == 0) {
      res++;
      currsum = 0;
      currN = i;
      i--;
    } else if (find(prevSum.begin() + currN, prevSum.end(), currsum) !=
               prevSum.end()) {
      res++;
      currsum = 0;
      currN = i;
      i--;
    } else {
      prevSum.push_back(currsum);
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    cout << 4 + 3 * n << endl;
    for (int x = 0; x <= n; x++) {
      cout << x << " " << x + 1 << endl;
    }
    for (int x = 0; x <= n + 1; x++) {
      cout << x << " " << x << endl;
    }
    for (int x = 0; x <= n; x++) {
      cout << x + 1 << " " << x << endl;
    }
  }
}
