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.
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≤1000) — 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 ui, vi (1≤ui,vi≤n) — meaning that there is an edge between vertices ui and vi. 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.
Meaning of the questions:
allows you to construct a map of n contacts, so that the total number of edges is a prime number, the degree of each node is also a prime number.
Ideas:
Properties using a n ~ n + n / 2 in this range, there must be a prime number.
FIG then we can put together into a ring, the number of edges now is n.
If the current number is not a prime number of sides, then starting from 1, 1 is added with its edges connected to each node in the ring opposite the 1 + n / 2 in the ring.
Prime number is not joined with the opposite side of node 2, so may be added up to n + n / 2 sides, depending on the nature of the above we can know, this process must sum the number of edges of a prime number.
Since this process has a degree of each node it is 2 or 3. Therefore, the overall condition is satisfied.
See details Code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
// const int maxn = 1e7+50;
bool noprime[maxn + 50];
vector <int> p;
void getPrime()
{
// 华丽的初始化
memset(noprime, false, sizeof(noprime));
p.clear();
int m = (int)sqrt(maxn + 0.5);
// 优化的埃筛
for (int i = 2; i <= m; i++)
{
if (!noprime[i])
{
for (int j = i * i; j <= maxn; j += i)
{
noprime[j] = true;
}
}
}
}
std::vector<pii> v;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
getPrime();
int n;
gbtb;
cin >> n;
if (!noprime[n])
{
cout << n << endl;
repd(i, 2, n)
{
cout << i << " " << i - 1 << endl;
}
cout << 1 << " " << n << endl;
} else
{
int ans = n;
int t = ans;
while (noprime[t])
{
t++;
}
cout << t << endl;
repd(i, 2, n)
{
cout << i << " " << i - 1 << endl;
}
cout << 1 << " " << n << endl;
int id = 1;
while (noprime[ans])
{
ans++;
cout << id << " " << id + n / 2 << endl;
id++;
}
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}