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: Accepted)

#include <bits/stdc++.h>

using namespace std;
#define local
#ifdef local

template<class T>
void _E(T x) { cerr << x; }

void _E(double x) { cerr << fixed << setprecision(6) << x; }

void _E(string s) { cerr << "\"" << s << "\""; }

template<class A, class B>
void _E(pair<A, B> x) {
    cerr << '(';
    _E(x.first);
    cerr << ", ";
    _E(x.second);
    cerr << ")";
}

template<class T>
void _E(vector<T> x) {
    cerr << "[";
    for (auto it = x.begin(); it != x.end(); ++it) {
        if (it != x.begin()) cerr << ", ";
        _E(*it);
    }
    cerr << "]";
}

void ERR() {}

template<class A, class... B>
void ERR(A x, B... y) {
    _E(x);
    cerr << (sizeof...(y) ? ", " : " ");
    ERR(y...);
}

#define debug(x...) do { cerr << "{ "#x" } -> { "; ERR(x); cerr << "}" << endl; } while(false)
#else
#define debug(...) 114514.1919810
#endif
#define _rep(n, a, b) for (int n = (a); n <= (b); ++n)
#define _rev(n, a, b) for (int n = (a); n >= (b); --n)
#define _for(n, a, b) for (int n = (a); n < (b); ++n)
#define _rof(n, a, b) for (int n = (a); n > (b); --n)
#define oo 0x3f3f3f3f3f3f3f3f
#define ll  long long
#define db long double
#define eps 1e-3
#define bin(x) cerr << bitset<10>(x) << endl
#define what_is(x) cerr << #x << " is " << x << endl
#define met(a, b) memset(a, b, sizeof(a))
#define all(x) x.begin(), x.end()
#define pii pair<ll, ll>
#define pdd pair<db, db>
#define endl "\n"
const ll mod = 998244353;
const ll maxn = 1e5 + 10;
const db pi = acos(-1.0);

ll qpow(ll a, ll b) {
    ll ret = 1;
    for (; b; a = a * a % mod, b >>= 1) {
        if (b & 1) {
            ret = ret * a % mod;
        }
    }
    return ret;
}

vector<ll> f(maxn), invf(maxn);

ll inv(ll a) {
    return qpow(a, mod - 2);
}

void prework() {
    f[0] = 1;
    _rep(i, 1, maxn - 1) {
        f[i] = f[i - 1] * i % mod;
    }
    invf[maxn - 1] = qpow(f[maxn - 1], mod - 2);
    for (ll i = maxn - 2; i >= 0; i--) {
        invf[i] = invf[i + 1] * (i + 1) % mod;
    }
}

ll C(ll n, ll m) {
    if (n > m || m < 0)return 0;
    if (n == 0 || m == n) return 1;
    ll res = (f[m] * invf[m - n] % mod * invf[n]) % mod;
    return res;
}

int par[maxn];

int find(int a) {
    return a == par[a] ? a : par[a] = find(par[a]);
}

void merge(int a, int b) {
    par[find(a)] = find(b);
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    _rep(i, 1, n) {
        par[i] = i;
    }
    _rep(i, 1, m) {
        int u, v;
        cin >> u >> v;
        merge(u, v);
    }
    int cnt = 0;
    _rep(i, 1, n) {
        if (par[i] == i) {
            cnt++;
        }
    }
    cout << cnt - 1 << endl;
}