Problem: 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.

Examples

Input

4 1
1 3


Output

2


Input

2 0


Output

0


Input

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


Output

12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)
#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)
#define all(v) (v).begin(), (v).end()
#define pb(q) push_back(q)
#define Abs(a,b) max(a,b)-min(a,b)
#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}
#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}
#define Cout(x) cout<<(x)<<endl
#define POSSIBLE(condition) if(condition){cout << "POSSIBLE" << endl;}else{cout << "IMPOSSIBLE" << endl;}
#define Possible(condition) if(condition){cout << "Possible" << endl;}else{cout << "Impossible" << endl;}
#define possible(condition) if(condition){cout << "possible" << endl;}else{cout << "impossible" << endl;}
#define Size(n) (n).size()

typedef long long ll;


using namespace std;

const int INF = 1e9,MOD = 1e9 + 7,ohara = 1e6;
const ll LINF = 1e18;


long long int n,cnt=0,ans=0,a,b,c,d,cmp,cmpp,m,h,w,x,y,sum=0,pos,par[ohara],ran[ohara];
int dy[]={1,0,-1,0};
int dx[]={0,1,0,-1};
string alph("abcdefghijklmnopqrstuvwxyz"),s;
bool fl=true;
struct edge{int to,cost;};

void init(int nn) {
	for (int i=0; i<nn; i++) {
		par[i] = i;
		ran[i] = 0;
	}
}//初期化

ll root(ll xx){
    if(par[xx]==xx){
        return xx;
    }
    else{
        return par[xx]=root(par[xx]);
    }
}
//木の根を求めている。
 
bool same(ll xx,ll yy){
    return root(xx)==root(yy);
}
//同じ集合？？
 
void unite(ll xx,ll yy){
    xx=root(xx);
    yy=root(yy);
    if(xx==yy)return;
    
    if(ran[xx]<ran[yy]){
        par[xx]=yy;
    }
    else{
        par[yy]=xx;
        if(ran[xx]==ran[yy])ran[xx]++;
    }
}
//xとyの属する集合を併合
 
//-------------------------↓↓↓↓↓↓------------------------

int main(void){
       cin.tie(0);
    ios::sync_with_stdio(false);


       cin>>n>>m;

       map<ll,ll> mp;

       init(n);

       rep(i,m){
          cin>>a>>b;
          unite(a,b);
       }
      
      rep(i,n){
        mp[root(i+1)]++;
      }
  
      cmp=mp[root(1)];
      cmpp=mp[root(2)];
      cmp=max(cmp,cmpp);

      rep(i,n){
        pos=mp[root(i+1)];
        if(root(i+1)!=root(1)&&root(i+1)!=root(2))cnt++;
        else{
        mp[root(i+1)]=0;
        rrep(i,1,pos){
          ans+=i;
        }
        }
      }

      rrep(i,1,cnt){
        ans+=i;
      }

      ans+=cmp*cnt;

      Cout(ans-m);
      
     

       
return 0;
}