Problem:
Score : 400 points

Problem Statement

There are N persons called Person 1 through Person N.

You are given M facts that "Person A_i and Person B_i are friends." The same fact may be given multiple times.

If X and Y are friends, and Y and Z are friends, then X and Z are also friends. There is no friendship that cannot be derived from the M given facts.

Takahashi the evil wants to divide the N persons into some number of groups so that every person has no friend in his/her group.

At least how many groups does he need to make?

Constraints

2 \leq N \leq 2\times 10^5

0 \leq M \leq 2\times 10^5

1\leq A_i,B_i\leq N

A_i \neq B_i

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
\vdots
A_M B_M

Output

Print the answer.

Sample Input 1

5 3
1 2
3 4
5 1

Sample Output 1

3

Dividing them into three groups such as \{1,3\}, \{2,4\}, and \{5\} achieves the goal.

Sample Input 2

4 10
1 2
2 1
1 2
2 1
1 2
1 3
1 4
2 3
2 4
3 4

Sample Output 2

4

Sample Input 3

10 4
3 1
4 1
5 9
2 6

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize ("O2")
#pragma GCC target ("avx2")
//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const int CM = 1 << 17, CL = 12;
char cn[CM + CL], * ci = cn + CM + CL, * owa = cn + CM, ct;
const ll ma0 = 1157442765409226768;
const ll ma1 = 1085102592571150095;
const ll ma2 = 71777214294589695;
const ll ma3 = 281470681808895;
const ll ma4 = 4294967295;
inline int getint() {
	if (ci - owa > 0) {
		memcpy(cn, owa, CL);
		ci -= CM;
		fread(cn + CL, 1, CM, stdin);
	}
	ll tmp = *(ll*)ci;
	int dig = 68 - __builtin_ctzll((tmp & ma0) ^ ma0);
	tmp = tmp << dig & ma1;
	tmp = tmp * 10 + (tmp >> 8) & ma2;
	tmp = tmp * 100 + (tmp >> 16) & ma3;
	tmp = tmp * 10000 + (tmp >> 32) & ma4;
	ci += 72 - dig >> 3;
	return tmp;
}
const int SZ = 200000;
class UF {
public:
	int P[SZ + 1], saidai;
	UF() : P(), saidai() {
		saidai = -1;
		rep1(i, SZ) P[i] = -1;
	}
	int find(int A) {
		if (P[A] < 0) return A;
		return P[A] = find(P[A]);
	}
	bool unite(int A, int B) {
		int a = find(A);
		int b = find(B);
		if (a == b) return false;
		if (P[a] > P[b]) swap(a, b);
		P[a] += P[b];
		if (P[a] < saidai) saidai = P[a];
		P[b] = a;
		return true;
	}
} uf;

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

	int N = getint(), M = getint();

	rep(i, M) {
		uf.unite(getint(), getint());
	}

	printf("%d\n", -uf.saidai);


	Would you please return 0;
}