#include <iostream>
using namespace std;
const int N=1e6+100;
int par[N];
int f(int n)
{
return (n*(n-1))/2;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int k;
cin>>k;
int n=0;
while(f(n+1)<=k)
{
n++;
}
k-=f(n);
// this will be atmost the sum of
for(int i=0;i<n;i++)
{
par[i]=i-1;
}
// 0 -1
// 1 0
int lst=n;
for(int i=n-1;i>=0;i--)
{
// if we add as child of i the number of ancestor are i+1
while(k>=(i+1))
{
k-=(i+1);
par[lst]=i;
lst++;
}
}
cout<<lst<<endl;
for(int i=0;i<lst;i++)cout<<i<<' '<<par[i]<<endl;;
// cout<<lst<<endl;
}
