Problem:
Score : 100 points

Problem Statement

You are given an undirected graph G. G has N vertices and M edges. The vertices are numbered from 1 through N, and the i-th edge (1 ≤ i ≤ M) connects Vertex a_i and b_i. G does not have self-loops and multiple edges.

You can repeatedly perform the operation of adding an edge between two vertices. However, G must not have self-loops or multiple edges as the result. Also, if Vertex 1 and 2 are connected directly or indirectly by edges, your body will be exposed to a voltage of 1000000007 volts. This must also be avoided.

Under these conditions, at most how many edges can you add? Note that Vertex 1 and 2 are never connected directly or indirectly in the beginning.

Constraints

2 ≤ N ≤ 10^5

0 ≤ M ≤ 10^5

1 ≤ a_i < b_i ≤ N

All pairs (a_i, b_i) are distinct.

Vertex 1 and 2 in G are not connected directly or indirectly by edges.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_M b_M

Output

Print the maximum number of edges that can be added.

Sample Input 1

4 1
1 3

Sample Output 1

2

As shown above, two edges can be added. It is not possible to add three or more edges.

Sample Input 2

2 0

Sample Output 2

0

No edge can be added.

Sample Input 3

9 6
1 4
1 8
2 5
3 6
4 8
5 7

Sample Output 3

12

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

typedef long long int int64;

int root(int *u,int x){
  if(u[x]==x) return x;
  return u[x]=root(u,u[x]);
}

int same(int *u,int x,int y){
  return root(u,x)==root(u,y);
}

void unite(int *u,int x,int y){
  x=root(u,x);
  y=root(u,y);
  if(x==y) return;
  u[x]=y;
}

void run(void){
  int n,m;
  scanf("%d%d",&n,&m);
  int *u=(int *)calloc(n+1,sizeof(int));
  int i;
  for(i=0;i<=n;i++) u[i]=i;
  for(i=0;i<m;i++){
    int a,b;
    scanf("%d%d",&a,&b);
    unite(u,a,b);
  }
  int a,b,c;
  a=b=c=0;
  for(i=1;i<=n;i++){
    if(same(u,1,i)){
      a++;
    } else if(same(u,2,i)){
      b++;
    } else {
      c++;
    }
  }
  if(a<=b){
    b+=c;
  } else {
    a+=c;
  }
  int64 ans=(int64)a*(a-1)/2+(int64)b*(b-1)/2-m;
  printf("%lld\n",ans);
}

int main(void){
  run();
  return 0;
}