Problem:
Score : 300 points

Problem Statement

There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M.
Road i connects City A_i and City B_i.

Snuke can perform the following operation zero or more times:

Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.

After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).

What is the minimum number of roads he must build to achieve the goal?

Constraints

2 \leq N \leq 100,000

1 \leq M \leq 100,000

1 \leq A_i < B_i \leq N

No two roads connect the same pair of cities.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

Print the answer.

Sample Input 1

3 1
1 2

Sample Output 1

1

Initially, there are three cities, and there is a road between City 1 and City 2.

Snuke can achieve the goal by building one new road, for example, between City 1 and City 3.
After that,

We can travel between 1 and 2 directly.

We can travel between 1 and 3 directly.

We can travel between 2 and 3 by following both roads (2 - 1 - 3).

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)


// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Graph class represents a undirected graph 
// using adjacency list representation 
class Graph { 
    // No. of vertices 
    int V; 
  
    // Pointer to an array containing adjacency lists 
    list<int>* adj; 
  
    // A function used by DFS 
    void DFSUtil(int v, bool visited[]); 
  
public: 
    // Constructor 
    Graph(int V); 
  
    void addEdge(int v, int w); 
    int NumberOfconnectedComponents(); 
}; 
  
// Function to return the number of 
// connected components in an undirected graph 
int Graph::NumberOfconnectedComponents() 
{ 
  
    // Mark all the vertices as not visited 
    bool* visited = new bool[V]; 
  
    // To store the number of connected components 
    int count = 0; 
    for (int v = 0; v < V; v++) 
        visited[v] = false; 
  
    for (int v = 0; v < V; v++) { 
        if (visited[v] == false) { 
            DFSUtil(v, visited); 
            count += 1; 
        } 
    } 
  
    return count; 
} 
  
void Graph::DFSUtil(int v, bool visited[]) 
{ 
  
    // Mark the current node as visited 
    visited[v] = true; 
  
    // Recur for all the vertices 
    // adjacent to this vertex 
    list<int>::iterator i; 
  
    for (i = adj[v].begin(); i != adj[v].end(); ++i) 
        if (!visited[*i]) 
            DFSUtil(*i, visited); 
} 
  
Graph::Graph(int V) 
{ 
    this->V = V; 
    adj = new list<int>[V]; 
} 
  
// Add an undirected edge 
void Graph::addEdge(int v, int w) 
{ 
    adj[v].push_back(w); 
    adj[w].push_back(v); 
} 
  
// Driver code 
int main() 
{ 
    int n,m,a,b;
    cin >> n >> m;
	Graph g(n);
	for(int i=0;i<m;i++)
	{ 
		cin >> a >> b;
    g.addEdge(a, b); 
}
  
    cout << g.NumberOfconnectedComponents()-1; 
  
    return 0; 
}