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 <iostream>
#include <iomanip> // << fixed << setprecision(xxx)
#include <algorithm> // do { } while ( next_permutation(A, A+xxx) ) ;
#include <vector>
#include <string> // to_string(nnn) // substr(m, n) // stoi(nnn)
#include <complex>
#include <tuple> // get<n>(xxx)
#include <queue>
#include <stack>
#include <map> // if (M.find(key) != M.end()) { }
#include <set> // S.insert(M);
// if (S.find(key) != S.end()) { }
// for (auto it=S.begin(); it != S.end(); it++) { }
// auto it = S.lower_bound(M);
#include <random> // random_device rd; mt19937 mt(rd());
#include <cctype>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib> // atoi(xxx)
using namespace std;

#define DEBUG 0 // change 0 -> 1 if we need debug.
// insert #if<tab> by my emacs. #if DEBUG == 1 ... #end

typedef long long ll;

// const int dx[4] = {1, 0, -1, 0};
// const int dy[4] = {0, 1, 0, -1};

// const int C = 1e6+10;
// const ll M = 1000000007;

ll hen(ll n) {
  return n * (n-1) / 2;
}

int main () {
  int N;
  cin >> N;
  int M;
  cin >> M;
  vector<int> V[100010];
  for (auto i = 0; i < M; ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    V[a].push_back(b);
    V[b].push_back(a);    
  }
  bool visited[100010];
  fill(visited, visited+100010, false);
  stack<int> S;
  S.push(0);
  int zero = 0;
  while (!S.empty()) {
    int src = S.top();
    S.pop();
    if (!visited[src]) {
      visited[src] = true;
      zero++;
      for (auto x : V[src]) {
        if (!visited[x]) {
          S.push(x);
        }
      }
    }
  }
  S.push(1);
  int one = 0;
  while (!S.empty()) {
    int src = S.top();
    S.pop();
    if (!visited[src]) {
      visited[src] = true;
      one++;
      for (auto x : V[src]) {
        if (!visited[x]) {
          S.push(x);
        }
      }
    }
  }
  if (one < zero) swap(one, zero);
  one = N - zero;
  cout << hen(one) + hen(zero) - M << endl;
}