#include <bits/stdc++.h>
const long long INF = (1LL << 60) - 1;
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long BINARY_SEARCH(long long dp[], long long n, long long key) {
  long long s = 1;
  long long e = n;
  while (s <= e) {
    long long mid = (s + e) / 2;
    if (dp[mid] == key)
      return mid;
    else if (dp[mid] > key)
      e = mid - 1;
    else
      s = mid + 1;
  }
  return -1;
}
string CONVERT_TO_BINARY(long long s) {
  string res = "";
  while (s != 0) {
    res += (char)('0' + s % 2);
    s /= 2;
  }
  reverse(res.begin(), res.end());
  return res;
}
bool PALIN(string s) {
  long long i = 0;
  long long j = s.length() - 1;
  while (i <= j) {
    if (s[i] != s[j]) return false;
    j--, i++;
  }
  return true;
}
long long STOI(string s) {
  long long num = 0;
  long long po = 1;
  for (long long i = s.length() - 1; i >= 0; i--) {
    num += po * (s[i] - '0');
    po *= 10;
  }
  return num;
}
long long modInverse(long long a, long long m) { return power(a, m - 2); }
int findLongestSub(string bin) {
  int n = bin.length(), i;
  int sum = 0;
  unordered_map<int, int> prevSum;
  int maxlen = 0;
  int currlen;
  for (i = 0; i < n; i++) {
    if (bin[i] == '1')
      sum++;
    else
      sum--;
    if (sum > 0) {
      maxlen = i + 1;
    } else if (sum <= 0) {
      if (prevSum.find(sum - 1) != prevSum.end()) {
        currlen = i - prevSum[sum - 1];
        maxlen = max(maxlen, currlen);
      }
    }
    if (prevSum.find(sum) == prevSum.end()) prevSum[sum] = i;
  }
  return maxlen;
}
vector<long long> prefix_KMP(string &s) {
  vector<long long> pi;
  pi[0] = 0;
  for (long long i = 1; i <= s.length() - 1; i++) {
    long long j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) {
      j = pi[j - 1];
    }
    if (s[i] == s[j]) {
      j++;
    }
    pi[i] = j;
  }
  return pi;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  cout << n / 2 + 1 << endl;
  ;
  long long cnto = 1;
  long long cnte = 1;
  for (long long i = 1; i <= n; i += 1) {
    if (i % 2 == 0) {
      cout << cnto << " " << cnte << endl;
      cnto++;
    } else {
      cout << cnto << " " << cnte << endl;
      cnte++;
    }
  }
}

### 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>
const long long INF = (1LL << 60) - 1;
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long BINARY_SEARCH(long long dp[], long long n, long long key) {
  long long s = 1;
  long long e = n;
  while (s <= e) {
    long long mid = (s + e) / 2;
    if (dp[mid] == key)
      return mid;
    else if (dp[mid] > key)
      e = mid - 1;
    else
      s = mid + 1;
  }
  return -1;
}
string CONVERT_TO_BINARY(long long s) {
  string res = "";
  while (s != 0) {
    res += (char)('0' + s % 2);
    s /= 2;
  }
  reverse(res.begin(), res.end());
  return res;
}
bool PALIN(string s) {
  long long i = 0;
  long long j = s.length() - 1;
  while (i <= j) {
    if (s[i] != s[j]) return false;
    j--, i++;
  }
  return true;
}
long long STOI(string s) {
  long long num = 0;
  long long po = 1;
  for (long long i = s.length() - 1; i >= 0; i--) {
    num += po * (s[i] - '0');
    po *= 10;
  }
  return num;
}
long long modInverse(long long a, long long m) { return power(a, m - 2); }
int findLongestSub(string bin) {
  int n = bin.length(), i;
  int sum = 0;
  unordered_map<int, int> prevSum;
  int maxlen = 0;
  int currlen;
  for (i = 0; i < n; i++) {
    if (bin[i] == '1')
      sum++;
    else
      sum--;
    if (sum > 0) {
      maxlen = i + 1;
    } else if (sum <= 0) {
      if (prevSum.find(sum - 1) != prevSum.end()) {
        currlen = i - prevSum[sum - 1];
        maxlen = max(maxlen, currlen);
      }
    }
    if (prevSum.find(sum) == prevSum.end()) prevSum[sum] = i;
  }
  return maxlen;
}
vector<long long> prefix_KMP(string &s) {
  vector<long long> pi;
  pi[0] = 0;
  for (long long i = 1; i <= s.length() - 1; i++) {
    long long j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) {
      j = pi[j - 1];
    }
    if (s[i] == s[j]) {
      j++;
    }
    pi[i] = j;
  }
  return pi;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  cout << n / 2 + 1 << endl;
  ;
  long long cnto = 1;
  long long cnte = 1;
  for (long long i = 1; i <= n; i += 1) {
    if (i % 2 == 0) {
      cout << cnto << " " << cnte << endl;
      cnto++;
    } else {
      cout << cnto << " " << cnte << endl;
      cnte++;
    }
  }
}

```