Problem:
Score : 700 points

Problem Statement

Takahashi is hosting an sports meet.
There are N people who will participate. These people are conveniently numbered 1 through N.
Also, there are M options of sports for this event. These sports are numbered 1 through M.
Among these options, Takahashi will select one or more sports (possibly all) to be played in the event.

Takahashi knows that Person i's j-th favorite sport is Sport A_{ij}.
Each person will only participate in his/her most favorite sport among the ones that are actually played in the event, and will not participate in the other sports.

Takahashi is worried that one of the sports will attract too many people.
Therefore, he would like to carefully select sports to be played so that the number of the participants in the sport with the largest number of participants is minimized.
Find the minimum possible number of the participants in the sport with the largest number of participants.

Constraints

1 \leq N \leq 300

1 \leq M \leq 300

A_{i1} , A_{i2} , ... , A_{iM} is a permutation of the integers from 1 to M.

Input

Input is given from Standard Input in the following format:

N M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
:
A_{N1} A_{N2} ... A_{NM}

Output

Print the minimum possible number of the participants in the sport with the largest number of participants.

Sample Input 1

4 5
5 1 3 4 2
2 5 3 1 4
2 3 1 4 5
2 5 4 3 1

Sample Output 1

2

Assume that Sports 1, 3 and 4 are selected to be played. In this case, Person 1 will participate in Sport 1, Person 2 in Sport 3, Person 3 in Sport 3 and Person 4 in Sport 4.
Here, the sport with the largest number of participants is Sport 3, with two participants.
There is no way to reduce the number of participants in the sport with the largest number of participants to 1. Therefore, the answer is 2.

Sample Input 2

3 3
2 1 3
2 1 3
2 1 3

Sample Output 2

3

Since all the people have the same taste in sports, there will be a sport with three participants, no matter what sports are selected.
Therefore, the answer is 3.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <queue>
#include <map>
#include <list>
#include <vector>
#include <string>
#include <stack>
#include <limits>
#include <climits>
#include <cassert>
#include <fstream>
#include <cstring>
#include <cmath>
#include <bitset>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <cstdio>
#include <ciso646>
#include <set>
#include <array>
#include <unordered_map>

using namespace std;

#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)

#define inf 0x3f3f3f3f
#define PB push_back
#define MP make_pair
#define ALL(a) (a).begin(),(a).end()
#define SET(a,c) memset(a,c,sizeof a)
#define CLR(a) memset(a,0,sizeof a)
#define VS vector<string>
#define VI vector<ll>
#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define MIN(a,b) (a>b?b:a)
#define MAX(a,b) (a>b?a:b)
#define pi 2*acos(0.0)
#define INFILE() freopen("in0.txt","r",stdin)
#define OUTFILE()freopen("out0.txt","w",stdout)
#define ll long long
#define ull unsigned long long
#define pii pair<ll,ll>
#define pcc pair<char,char>
#define pic pair<ll,char>
#define pci pair<char,ll>
#define eps 1e-14
#define FST first
#define SEC second
#define SETUP cin.tie(0), ios::sync_with_stdio(false), cout << setprecision(15)

namespace {
	struct input_returnner {
		ll N; input_returnner(ll N_ = 0) :N(N_) {}
		template<typename T> operator vector<T>() const { vector<T> res(N); for (auto &a : res) cin >> a; return std::move(res); }
		template<typename T> operator T() const { T res; cin >> res; return res; }
		template<typename T> T operator - (T right) { return T(input_returnner()) - right; }
		template<typename T> T operator + (T right) { return T(input_returnner()) + right; }
		template<typename T> T operator * (T right) { return T(input_returnner()) * right; }
		template<typename T> T operator / (T right) { return T(input_returnner()) / right; }
		template<typename T> T operator << (T right) { return T(input_returnner()) << right; }
		template<typename T> T operator >> (T right) { return T(input_returnner()) >> right; }
	};
	template<typename T> input_returnner in() { return in<T>(); }
	input_returnner in() { return input_returnner(); }
	input_returnner in(ll N) { return std::move(input_returnner(N)); }
}

const ll MOD = 1e9 + 7;

struct ModInt {
	ll v = 0;
	ModInt() {}
	template<class T> ModInt(const T& right) {
		v = right;
		if (v >= 0) v %= MOD;
		else v += ((-v) / MOD + 1)*MOD;
		v %= MOD;
	}
	void operator = (const ModInt& right) { v = right.v; }
	template<class T> void operator = (const T& right) {
		v = right;
		if (v >= 0) v %= MOD;
		else v = v += ((-v) / MOD + 1)*MOD;
		v %= MOD;
	}

	ModInt operator + (const ModInt& right) { return (v + right.v) % MOD; }
	ModInt operator - (const ModInt& right) { return (MOD - (v - right.v)); }
	ModInt operator * (const ModInt& right) { return (v * right.v) % MOD; }
	ModInt operator / (const ModInt& right) { return v / right.v; }

	void operator += (const ModInt& right) { v = (v + right.v) % MOD; }
	void operator -= (const ModInt& right) { v = (MOD - (v - right.v)); }
	void operator *= (const ModInt& right) { v = (v* right.v) % MOD; }
	void operator /= (const ModInt& right) { v = v / right.v; }

	bool operator == (const ModInt& right) { return v == right.v; }
};

ostream& operator << (ostream& os, const ModInt& value) {
	os << value.v;
	return os;
}

string YN[] = { "NO", "YES" };

int gcd ( int a, int b )
{
  if ( a==0 ) return b;
  return gcd ( b%a, a );
}



void solve();
/// ---template---

signed main(void) {
	SETUP;
	solve();
#ifdef _DEBUG
	system("pause");
#endif
	return 0;
}

int A[300][300] = {};
int N, M;

//checked: もう使ってはならない
//cnt: 各競技の参加人数 numを越えてはならない
//depth: 何人目を見ているか
//selected: どの競技を選択したのか
//return: ダメだった番号
//retrun: -1 OK
//retrun: -100 NG
int check(int num, vector<int>& cnt, vector<bool>& checked, int depth) {
	if (depth == N) return -1;
	REP(j, M) {
		if (checked[A[depth][j]]) continue;
		if (cnt[A[depth][j]] < num) {
			// cerr << "num(" << num << "), " << "depth(" << depth << "), " << "check:" << A[depth][j] << endl;
			cnt[A[depth][j]]++;
		}
		else {
			checked[A[depth][j]] = true;
			return A[depth][j];
		}

		int c = check(num, cnt, checked, depth + 1);
		if (c == -1) {
			return -1;
		}
		//違ってた場合はダメだったやつが消えるまで遡る
		else {
			cnt[A[depth][j]]--;
			// cerr << "num(" << num << "), " << "depth(" << depth << "), " << "delete:" << A[depth][j] << endl;
			if (cnt[c] > 0) return c;
		}
	}
	return -100;
}

void solve() {
	cin >> N >> M;
	REP(i, N) REP(j, M) cin >> A[i][j];

	int high = N;
	int low = 0;
	int mid;
	while ( high - low > 1) {
		mid = (low + high) / 2;
		vector<int> cnt(M+1);
		vector<bool> checked(M+1);
		if (check(mid, cnt, checked, 0) == -1) high = mid;
		else low = mid;
	}
	cout << high << endl;
}