Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!

When building the graph, he needs four conditions to be satisfied: 

  * It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops. 
  * The number of vertices must be exactly n — a number he selected. This number is not necessarily prime. 
  * The total number of edges must be prime. 
  * The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime. 



Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4. 

<image>

Note that the graph can be disconnected.

Please help Bob to find any such graph!

Input

The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.

Output

If there is no graph satisfying the conditions, print a single line containing the integer -1.

Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.

If there are multiple solutions, you may print any of them.

Note that the graph can be disconnected.

Examples

Input


4


Output


5
1 2
1 3
2 3
2 4
3 4

Input


8


Output


13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8

Note

The first example was described in the statement.

In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.

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#include <bits/stdc++.h>
bool debug = true;
using namespace std;
static auto start = chrono::high_resolution_clock::now();
int prime(int n) {
  int i, j;
  for (i = 2; i * i <= n; i++) {
    if (!(n % i)) return 0;
  }
  return 1;
}
int main(void) {
  if (0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  int i, j, k, n, m, mx;
  scanf("%d", &n);
  m = n;
  mx = (n * (n - 1)) >> 1;
  for (; m <= mx; m++) {
    if (prime(m)) break;
  }
  printf("%d\n", m);
  for (i = 0; i < n; i++) {
    printf("%d %d\n", i + 1, ((i + 1) % n) + 1);
  }
  m -= n;
  vector<int> vis(n);
  for (i = 0; i < n - 2; i++) {
    if (!m) break;
    if (!vis[i]) {
      vis[i + 2] = true;
      printf("%d %d\n", i + 1, i + 3);
      m--;
    }
  }
  auto stop = chrono::high_resolution_clock::now();
  auto duration = chrono::duration_cast<chrono::microseconds>(stop - start);
  if (0) {
    printf("Time in seconds: %.6lf\n", (double)duration.count() / 1000000);
  }
}
