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:
//start of jonathanirvings' template v3.0.3 (BETA)

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <deque>
#include <string>
#include <string.h>
#include <vector>
#include <stack>
#include <queue>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <time.h>
#include <bitset>
#include <list>
#include <assert.h>
#include <time.h>
using namespace std;

typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
typedef pair<string,string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<LL> vl;
typedef vector<vl> vvl;

double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1,0,0,1,-1,-1,1,1};
int diry[8] = {0,1,-1,0,-1,1,-1,1};

#ifdef TESTING
  #define DEBUG fprintf(stderr,"====TESTING====\n")
  #define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
  #define debug(...) fprintf(stderr, __VA_ARGS__)
#else
  #define DEBUG 
  #define VALUE(x)
  #define debug(...)
#endif

#define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a))
#define FORD(a,b,c) for (int (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define REP(i,n) FOR(i,0,n)
#define REPN(i,n) FORN(i,1,n)
#define MAX(a,b) a = max(a,b)
#define MIN(a,b) a = min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define RESET(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)

inline string IntToString(LL a){
  char x[100];
  sprintf(x,"%lld",a); string s = x;
  return s;
}

inline LL StringToInt(string a){
  char x[100]; LL res;
  strcpy(x,a.c_str()); sscanf(x,"%lld",&res);
  return res;
}

inline string GetString(void){
  char x[1000005];
  scanf("%s",x); string s = x;
  return s;
}

inline string uppercase(string s){
  int n = SIZE(s); 
  REP(i,n) if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}

inline string lowercase(string s){
  int n = SIZE(s); 
  REP(i,n) if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}

inline void OPEN (string s) {
  #ifndef TESTING
  freopen ((s + ".in").c_str (), "r", stdin);
  freopen ((s + ".out").c_str (), "w", stdout);
  #endif
}

//end of jonathanirvings' template v3.0.3 (BETA)

int n,m;
vi adj[100005];
int a,b;
bool visited[100005];

LL C(int x)
{
  return (LL)x * (x - 1) / 2;
}

int dfs(int x)
{
  if (visited[x]) return 0;
  visited[x] = 1;
  int ret = 1;
  REP(i,SIZE(adj[x]))
  {
    ret += dfs(adj[x][i]);
  }
  return ret;
}

int main()
{
  scanf("%d %d",&n,&m);
  REP(i,m)
  {
    scanf("%d %d",&a,&b);
    adj[a].pb(b);
    adj[b].pb(a);
  }
  a = dfs(1);
  b = dfs(2);
  // debug("%d %d\n",a,b);
  if (a < b) swap(a,b);
  LL risan = C(n - b) + C(b) - m;
  printf("%lld\n",risan);
  return 0;
}