Problem:
Problem B: The Sorcerer's Donut

Your master went to the town for a day. You could have a relaxed day without hearing his
scolding. But he ordered you to make donuts dough by the evening. Loving donuts so much, he
can't live without eating tens of donuts everyday. What a chore for such a beautiful day.

But last week, you overheard a magic spell that your master was using. It was the time to
try. You casted the spell on a broomstick sitting on a corner of the kitchen. With a flash of
lights, the broom sprouted two arms and two legs, and became alive. You ordered him, then he
brought flour from the storage, and started kneading dough. The spell worked, and how fast he
kneaded it!

A few minutes later, there was a tall pile of dough on the kitchen table. That was enough for
the next week. \OK, stop now." You ordered. But he didn't stop. Help! You didn't know the
spell to stop him! Soon the kitchen table was filled with hundreds of pieces of dough, and he
still worked as fast as he could. If you could not stop him now, you would be choked in the
kitchen filled with pieces of dough.

Wait, didn't your master write his spells on his notebooks? You went to his den, and found the
notebook that recorded the spell of cessation.

But it was not the end of the story. The spell written in the notebook is not easily read by
others. He used a plastic model of a donut as a notebook for recording the spell. He split the
surface of the donut-shaped model into square mesh (Figure B.1), and filled with the letters
(Figure B.2). He hid the spell so carefully that the pattern on the surface looked meaningless.
But you knew that he wrote the pattern so that the spell "appears" more than once (see the next
paragraph for the precise conditions). The spell was not necessarily written in the left-to-right
direction, but any of the 8 directions, namely left-to-right, right-to-left, top-down, bottom-up,
and the 4 diagonal directions.

You should be able to find the spell as the longest string that appears more than once. Here,
a string is considered to appear more than once if there are square sequences having the string
on the donut that satisfy the following conditions.

Each square sequence does not overlap itself. (Two square sequences can share some squares.)

The square sequences start from different squares, and/or go to different directions.

Figure B.1: The Sorcerer's Donut Before
Filled with Letters, Showing the Mesh and 8
Possible Spell Directions

Figure B.2: The Sorcerer's Donut After Filled
with Letters

Note that a palindrome (i.e., a string that is the same whether you read it backwards or forwards)
that satisfies the first condition "appears" twice.

The pattern on the donut is given as a matrix of letters as follows.

ABCD
EFGH
IJKL

Note that the surface of the donut has no ends; the top and bottom sides, and the left and right
sides of the pattern are respectively connected. There can be square sequences longer than both
the vertical and horizontal lengths of the pattern. For example, from the letter F in the above
pattern, the strings in the longest non-self-overlapping sequences towards the 8 directions are
as follows.

FGHE
FKDEJCHIBGLA
FJB
FIDGJAHKBELC
FEHG
FALGBIHCJEDK
FBJ
FCLEBKHAJGDI

Please write a program that finds the magic spell before you will be choked with pieces of donuts
dough.

Input

The input is a sequence of datasets. Each dataset begins with a line of two integers h and w,
which denote the size of the pattern, followed by h lines of w uppercase letters from A to Z,
inclusive, which denote the pattern on the donut. You may assume 3 ≤ h ≤ 10 and 3 ≤ w ≤ 20.

The end of the input is indicated by a line containing two zeros.

Output

For each dataset, output the magic spell. If there is more than one longest string of the same
length, the first one in the dictionary order must be the spell. The spell is known to be at least
two letters long. When no spell is found, output 0 (zero).

Sample Input

5 7
RRCABXT
AABMFAB
RROMJAC
APTADAB
YABADAO
3 13
ABCDEFGHIJKLM
XMADAMIMADAMY
ACEGIKMOQSUWY
3 4
DEFG
ACAB
HIJK
3 6
ABCDEF
GHIAKL
MNOPQR
10 19
JFZODYDXMZZPEYTRNCW
XVGHPOKEYNZTQFZJKOD
EYEHHQKHFZOVNRGOOLP
QFZOIHRQMGHPNISHXOC
DRGILJHSQEHHQLYTILL
NCSHQMKHTZZIHRPAUJA
NCCTINCLAUTFJHSZBVK
LPBAUJIUMBVQYKHTZCW
XMYHBVKUGNCWTLLAUID
EYNDCCWLEOODXYUMBVN
0 0

Output for the Sample Input

ABRACADABRA
MADAMIMADAM
ABAC
0
ABCDEFGHIJKLMNOPQRSTUVWXYZHHHHHABCDEFGHIJKLMNOPQRSTUVWXYZ

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

#include<bits/stdc++.h>

using namespace std;
#define INFS (1LL<<28)
#define DEKAI 1000000007
#define INF 1000000007
//#define MOD 1000000007
#define lp(i,n) for(int i=0;i<n;i++)
#define lps(i,n) for(int i=1;i<=n;i++)
#define all(c) begin(c), end(c)

//#define int long long 

namespace {
	#define __DECLARE__(C)    \
	    template <typename T> \
	std::ostream &operator<<(std::ostream &, const C<T> &);

	#define __DECLAREM__(C)               \
	    template <typename T, typename U> \
	std::ostream &operator<<(std::ostream &, const C<T, U> &);

	__DECLARE__(std::vector)
	__DECLARE__(std::deque)
	__DECLARE__(std::set)
	__DECLARE__(std::stack)
	__DECLARE__(std::queue)
	__DECLARE__(std::priority_queue)
	__DECLARE__(std::unordered_set)
	__DECLAREM__(std::map)
	__DECLAREM__(std::unordered_map)

	template <typename T, typename U>
	std::ostream &operator<<(std::ostream &, const std::pair<T, U> &);
	template <typename... T>
	std::ostream &operator<<(std::ostream &, const std::tuple<T...> &);
	template <typename T, std::size_t N>
	std::ostream &operator<<(std::ostream &, const std::array<T, N> &);

	template <typename Tuple, std::size_t N>
	struct __TuplePrinter__ {
		static void print(std::ostream &os, const Tuple &t) {
			__TuplePrinter__<Tuple, N - 1>::print(os, t);
			os << ", " << std::get<N - 1>(t);
		}
	};

	template <typename Tuple>
	struct __TuplePrinter__<Tuple, 1> {
		static void print(std::ostream &os, const Tuple &t) { os << std::get<0>(t); }
	};

	template <typename... T>
	std::ostream &operator<<(std::ostream &os, const std::tuple<T...> &t) {
		os << '(';
		__TuplePrinter__<decltype(t), sizeof...(T)>::print(os, t);
		os << ')';
		return os;
	}

	template <typename T, typename U>
	std::ostream &operator<<(std::ostream &os, const std::pair<T, U> &v) {
		return os << '(' << v.first << ", " << v.second << ')';
	}

	#define __INNER__                             \
	os << '[';                                \
	for (auto it = begin(c); it != end(c);) { \
		os << *it;                            \
		os << (++it != end(c) ? ", " : "");   \
	}                                         \
	return os << ']';

	template <typename T, std::size_t N>
	std::ostream &operator<<(std::ostream &os, const std::array<T, N> &c) {
		__INNER__
	}

	#define __DEFINE__(C)                                           \
	    template <typename T>                                       \
	std::ostream &operator<<(std::ostream &os, const C<T> &c) { \
		__INNER__                                               \
	}

	#define __DEFINEM__(C)                                             \
	    template <typename T, typename U>                              \
	std::ostream &operator<<(std::ostream &os, const C<T, U> &c) { \
		__INNER__                                                  \
	}

	#define __DEFINEW__(C, M1, M2)                                  \
	    template <typename T>                                       \
	std::ostream &operator<<(std::ostream &os, const C<T> &c) { \
		std::deque<T> v;                                        \
		for (auto d = c; !d.empty(); d.pop()) v.M1(d.M2());     \
			return os << v;                                         \
	}

	__DEFINE__(std::vector)
	__DEFINE__(std::deque)
	__DEFINE__(std::set)
	__DEFINEW__(std::stack, push_front, top)
	__DEFINEW__(std::queue, push_back, front)
	__DEFINEW__(std::priority_queue, push_front, top)
	__DEFINE__(std::unordered_set)
	__DEFINEM__(std::map)
	__DEFINEM__(std::unordered_map)
}

#define pii pair<int,int>
#define ll long long
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

// modint 
template <signed M, unsigned T>
struct mod_int {
	constexpr static signed MODULO = M;
	constexpr static unsigned TABLE_SIZE = T;

	signed x;

	mod_int() : x(0) {}

	mod_int(long long y) : x(static_cast<signed>(y >= 0 ? y % MODULO : MODULO - (-y) % MODULO)) {}

	mod_int(int y) : x(y >= 0 ? y % MODULO : MODULO - (-y) % MODULO) {}

	mod_int &operator+=(const mod_int &rhs) {
		if ((x += rhs.x) >= MODULO) x -= MODULO;
		return *this;
	}

	mod_int &operator-=(const mod_int &rhs) {
		if ((x += MODULO - rhs.x) >= MODULO) x -= MODULO;
		return *this;
	}

	mod_int &operator*=(const mod_int &rhs) {
		x = static_cast<signed>(1LL * x * rhs.x % MODULO);
		return *this;
	}

	mod_int &operator/=(const mod_int &rhs) {
		x = static_cast<signed>((1LL * x * rhs.inv().x) % MODULO);
		return *this;
	}

	mod_int operator-() const { return mod_int(-x); }

	mod_int operator+(const mod_int &rhs) const { return mod_int(*this) += rhs; }

	mod_int operator-(const mod_int &rhs) const { return mod_int(*this) -= rhs; }

	mod_int operator*(const mod_int &rhs) const { return mod_int(*this) *= rhs; }

	mod_int operator/(const mod_int &rhs) const { return mod_int(*this) /= rhs; }

	bool operator<(const mod_int &rhs) const { return x < rhs.x; }

	mod_int inv() const {
		assert(x != 0);
		if (x <= static_cast<signed>(TABLE_SIZE)) {
			if (_inv[1].x == 0) prepare();
			return _inv[x];
		} else {
			signed a = x, b = MODULO, u = 1, v = 0, t;
			while (b) {
				t = a / b;
				a -= t * b;
				std::swap(a, b);
				u -= t * v;
				std::swap(u, v);
			}
			return mod_int(u);
		}
	}

	mod_int pow(long long t) const {
		assert(!(x == 0 && t == 0));
		mod_int e = *this, res = mod_int(1);
		for (; t; e *= e, t >>= 1)
			if (t & 1) res *= e;
		return res;
	}

	mod_int fact() {
		if (_fact[0].x == 0) prepare();
		return _fact[x];
	}

	mod_int inv_fact() {
		if (_fact[0].x == 0) prepare();
		return _inv_fact[x];
	}

	mod_int choose(mod_int y) {
		assert(y.x <= x);
		return this->fact() * y.inv_fact() * mod_int(x - y.x).inv_fact();
	}

	static mod_int _inv[TABLE_SIZE + 1];

	static mod_int _fact[TABLE_SIZE + 1];

	static mod_int _inv_fact[TABLE_SIZE + 1];

	static void prepare() {
		_inv[1] = 1;
		for (int i = 2; i <= (int)TABLE_SIZE; ++i) {
			_inv[i] = 1LL * _inv[MODULO % i].x * (MODULO - MODULO / i) % MODULO;
		}
		_fact[0] = 1;
		for (unsigned i = 1; i <= TABLE_SIZE; ++i) {
			_fact[i] = _fact[i - 1] * int(i);
		}
		_inv_fact[TABLE_SIZE] = _fact[TABLE_SIZE].inv();
		for (int i = (int)TABLE_SIZE - 1; i >= 0; --i) {
			_inv_fact[i] = _inv_fact[i + 1] * (i + 1);
		}
	}
};

template <int M, unsigned F>
std::ostream &operator<<(std::ostream &os, const mod_int<M, F> &rhs) {
	return os << rhs.x;
}

template <int M, unsigned F>
std::istream &operator>>(std::istream &is, mod_int<M, F> &rhs) {
	long long s;
	is >> s;
	rhs = mod_int<M, F>(s);
	return is;
}

template <int M, unsigned F>
mod_int<M, F> mod_int<M, F>::_inv[TABLE_SIZE + 1];

template <int M, unsigned F>
mod_int<M, F> mod_int<M, F>::_fact[TABLE_SIZE + 1];

template <int M, unsigned F>
mod_int<M, F> mod_int<M, F>::_inv_fact[TABLE_SIZE + 1];

template <int M, unsigned F>
bool operator==(const mod_int<M, F> &lhs, const mod_int<M, F> &rhs) {
	return lhs.x == rhs.x;
}

template <int M, unsigned F>
bool operator!=(const mod_int<M, F> &lhs, const mod_int<M, F> &rhs) {
	return !(lhs == rhs);
}

const int MF = 1000010;
const int MOD = 1000000007;

using mint = mod_int<MOD, MF>;

mint binom(int n, int r) { return (r < 0 || r > n || n < 0) ? 0 : mint(n).choose(r); }

mint fact(int n) { return mint(n).fact(); }

mint inv_fact(int n) { return mint(n).inv_fact(); }
const ll mod = 1000000007;
const int MAX_N = 10000;  // 400MB
// const int MAX_N = 1024; // 4MB
// nCr % mod

#define int long long
#define double long double 

inline ll gcds(ll a, ll b) { return b ? gcds(b, a % b) : a; }
inline ll lcms(ll a, ll b) { return a / gcd(a, b) * b; }

#define RK 200000000000
#define LK 300000000000
#define PL 400000000000
#define MI 500000000000
#define KA 600000000000

string solve(vector<string> v){
	set<string> ls;
	int ans=-1;
	string as="0";
	lp(i,v.size()){
		string s=v[i]+v[i];
		//cout<<s<<endl;
		lp(len,v[i].size()+1){
			if(len==0)continue;
			for(int j=0;j<v[i].size();j++){
				string t=s.substr(j,len);
				if(ls.find(t)!=ls.end()){
					if(ans<(int)t.size()){
						as=t;
						ans=t.size();
					}
					else if (ans==(int)t.size()&&as>t){
						as=t;
						ans=t.size();
					}/*
					else if (ans==t.size()){
						cout<<as<<" "<<t<<endl;
					}*/
				}
				else{
					ls.insert(t);
				}
				//cout<<t<<endl;
			}
		}
	}
	//cout<<"A:"<<as<<endl;
	if(as.size()<=1)as="0";
	if(ans==-1||ans==1)as="0";
	return as;
}

signed main(){
	while(1){
		int h,w;
		cin>>h>>w;
		if(h==0&&w==0)break;
		vector<string> s(h);
		vector<string> lis;
		lp(i,h){
			cin>>s[i];
			lis.push_back(s[i]);
			string t=s[i];
			reverse(all(t));
			lis.push_back(t);
		}
		lp(i,w){
			string t="";
			lp(j,h){
				t+=s[j][i];
			}
			lis.push_back(t);
			reverse(all(t));
			lis.push_back(t);
		}
		int done[20];
		lp(i,20)done[i]=0;
		lp(z,h){
			if(done[z]==1)break;
			int nowy=0,nowx=z;
			string t="";
			while(1){//migisita
				t+=s[nowy][nowx];
				if(nowy==0)done[nowx]=1;
				nowx+=1;
				nowy+=1;
				nowx%=w;
				nowy%=h;
				if(nowy==0&&nowx==z)break;
			}
			lis.push_back(t);
			reverse(all(t));
			lis.push_back(t);

			nowy=0,nowx=z;
			t="";
			while(1){//migisita
				t+=s[nowy][nowx];
				nowx+=1;
				nowy+=-1;
				nowx%=w;
				nowy+=h;
				(nowy)%=h;
				if(nowy==0&&nowx==z)break;
			}
			lis.push_back(t);
			reverse(all(t));
			lis.push_back(t);
		}
		//cout<<lis<<endl;
		cout<<solve(lis)<<endl;
	}
	return 0;
}