Problem:
Score : 1900 points

Problem Statement

Snuke has a grid with H rows and W columns. The square at the i-th row and j-th column contains a character S_{i,j}.

He can perform the following two kinds of operation on the grid:

Row-reverse: Reverse the order of the squares in a selected row.

Column-reverse: Reverse the order of the squares in a selected column.

For example, reversing the 2-nd row followed by reversing the 4-th column will result as follows:

By performing these operations any number of times in any order, how many placements of characters on the grid can be obtained?

Constraints

1≦H,W≦200

S_{i,j} is a lowercase English letter (a-z).

Input

The input is given from Standard Input in the following format:

H W
S_{1,1}S_{1,2}...S_{1,W}
S_{2,1}S_{2,2}...S_{2,W}
:
S_{H,1}S_{H,2}...S_{H,W}

Output

Print the number of placements of characters on the grid that can be obtained, modulo 1000000007 (=10^9+7).

Sample Input 1

2 2
cf
cf

Sample Output 1

6

The following 6 placements of characters can be obtained:

Sample Input 2

1 12
codefestival

Sample Output 2

2

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


//#define USEPB_DS
#define USETR1
#define CPPELEVEN
#define GPP

/*
 * temp.cpp
 *
 *  Created on: 2012-7-18
 *      Author: BSBandme
 */
//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <iostream>
#include <fstream>
#include <string.h>
#include <cstdio>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <cassert>
#include <list>
#include <iomanip>
#include <math.h>
#include <deque>
#include <utility>
#include <map>
#include <set>
#include <bitset>
#include <numeric>
#include <climits>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <functional>
#include <sstream>
#include <tuple>

using namespace std;

#ifndef CPPELEVEN
#ifdef USETR1
#include <tr1/unordered_map>
#include <tr1/unordered_set>
using namespace tr1;
#endif
#else
#include <unordered_map>
#include <unordered_set>
#endif

#ifdef USEPB_DS
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
// binomial_heap_tag, rc_binomial_heap_tag, thin_heap_tag, binary_heap_tag
typedef __gnu_pbds::priority_queue<int, greater<int>, pairing_heap_tag> pq_type;
// splay_tree_tag, ov_tree_tag
typedef tree <int, null_type, less <int>, rb_tree_tag, tree_order_statistics_node_update> tree_type;
#endif

#define mpr make_pair
typedef unsigned int ui;
typedef unsigned long long ull;
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <double, double> pdd;
typedef vector <int> vi;
typedef vector <ll> vll;
typedef vector <double> vd;
typedef vector <string> vs;
typedef map <string, int> mpsi;
typedef map <double, int> mpdi;
typedef map <int, int> mpii;

const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};

template <class T> inline T abs1(T a) {return a < 0 ? -a : a;}

#ifndef CPPELEVEN
template <class T> inline T max1(T a, T b) { return b < a ? a : b; }
template <class T> inline T max1(T a, T b, T c) { return max1(max1(a, b), c); }
template <class T> inline T max1(T a, T b, T c, T d) { return max1(max1(a, b, c), d); }
template <class T> inline T max1(T a, T b, T c, T d, T e) { return max1(max1(a, b, c, d), e); }
template <class T> inline T min1(T a, T b) { return a < b ? a : b; }
template <class T> inline T min1(T a, T b, T c) { return min1(min1(a, b), c); }
template <class T> inline T min1(T a, T b, T c, T d) { return min1(min1(a, b, c), d); }
template <class T> inline T min1(T a, T b, T c, T d, T e) { return min1(min1(a, b, c, d), e); }
#else
template <typename t, typename t1>
t min1(t a, t1 b) { return a < b ? a : b; }
template <typename t, typename... arg>
t min1(t a, arg... arr) { return min1(a, min1(arr...)); }
template <typename t, typename t1>
t max1(t a, t1 b) { return a > b ? a : b; }
template <typename t, typename... arg>
t max1(t a, arg... arr) { return max1(a, max1(arr...)); }
#endif

inline int jud(double a, double b){
	if(abs(a) < eps && abs(b) < eps) return 0;
	else if(abs1(a - b) / max(abs1(a), abs1(b)) < eps) return 0;
	if(a < b) return -1;
	return 1;
}
template <typename t> inline int jud(t a, t b){
	if(a < b) return -1;
	if(a == b) return 0;
	return 1;
}

// f_lb == 1代表返回相同的一串的左边界，f_small == 1代表返回如果没有寻找的值返回小的数
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1){
	if(na == 0) return 0;
	int be = 0, en = na - 1;
	if(*a <= *(a + na - 1)){
		if(f_lb == 0) while(be < en){
			int mid = (be + en + 1) / 2;
			if(jud(*(a + mid), val) != 1) be = mid;
			else en = mid - 1;
		}else while(be < en){
			int mid = (be + en) / 2;
			if(jud(*(a + mid), val) != -1) en = mid;
			else be = mid + 1;
		}
		if(f_small && jud(*(a + be), val) == 1) be--;
		if(!f_small && jud(*(a + be), val) == -1) be++;
	} else {
		if(f_lb) while(be < en){
			int mid = (be + en + 1) / 2;
			if(jud(*(a + mid), val) != -1) be = mid;
			else en = mid - 1;
		}else while(be < en){
			int mid = (be + en) / 2;
			if(jud(*(a + mid), val) != 1) en = mid;
			else be = mid + 1;
		}
		if(!f_small && jud(*(a + be), val) == -1) be--;
		if(f_small && jud(*(a + be), val) == 1) be++;
	}
	return be;
}

template <class T> inline T lowb(T num) {return num & (-num); }
#ifdef GPP
inline int bitnum(ui nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(ull nValue) { return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32); }
inline int bitnum(ll nValue) { return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32); }
inline int bitmaxl(ui a) { if(a == 0) return 0; return 32 - __builtin_clz(a); }
inline int bitmaxl(int a) { if(a == 0) return 0; return 32 - __builtin_clz(a); }
inline int bitmaxl(ull a) { int temp = a >> 32; if(temp) return 32 - __builtin_clz(temp) + 32; return bitmaxl(int(a)); }
inline int bitmaxl(ll a) { int temp = a >> 32; if(temp) return 32 - __builtin_clz(temp) + 32; return bitmaxl(int(a)); }
#else
#endif

long long pow(long long n, long long m, long long mod = 0){
	if(m < 0) return 0;
	long long ans = 1;
	long long k = n;
	while(m){
		if(m & 1) {
			ans *= k;
			if(mod) ans %= mod;
		}
		k *= k;
		if(mod) k %= mod;
		m >>= 1;
	}
	return ans;
}

#define  MOD 1000000007
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
	if(mod == -1) mod = MOD;
	a += b;
	while(a >= mod) a -= mod;
	while(a < 0) a += mod;
}
template <class t>
void output1(t arr) {
	for(int i = 0; i < (int)arr.size(); i++)
		cerr << arr[i] << ' ';
	cerr << endl;
}
template <class t>
void output2(t arr) {
	for(int i = 0; i < (int)arr.size(); i++)
		output1(arr[i]);
}

//....................密..........封..........线..........下..........禁..........止..........hack...............................................

const int mod = MOD;
const int maxn = 210;
int n, m;
char mp[maxn][maxn];

int main() {

//............................不要再忘了检查maxn大小了！！！！BSBandme你个SB！！！！...................................................

	ios_base::sync_with_stdio(0);
	#ifdef DEBUG //......................................................................................................
	freopen("input.txt", "r", stdin);
	#endif //...........................................................................................................

	scanf("%d%d", &n, &m);
	for(int i = 0 ;i < n; i++)
		scanf("%s", mp[i]);
	ll ans = 1;
	for(int i = 0, li = n - 1; i < li; i++, li--) {
		for(int j = 0, lj = m - 1; j < lj; j++, lj--) {
			map <char, int> stat;
			stat[mp[i][j]]++;
			stat[mp[li][j]]++;
			stat[mp[i][lj]]++;
			stat[mp[li][lj]]++;
			vi rs;
			for(auto it : stat)
				rs.push_back(it.second);
			sort(rs.begin(), rs.end());
			if(rs == vi{1,1,1,1}) ans *= 24;
			if(rs == vi{1,1,2}) ans *= 12;
			if(rs == vi{1,3}) ans *= 4;
			if(rs == vi{2,2}) ans *= 6;
			if(rs == vi{4}) ans *= 1;
			ans %= mod;
		}
	}
	int mult = 1;
	if(n % 2 == 1) {
		for(int i = 0, li = m - 1; i < li; i++)
			if(mp[n / 2][i] != mp[n / 2][li])
				mult = 2;
	}
	ans *= mult;
	ans %= mod;
	mult = 1;
	if(m % 2 == 1) {
		for(int i = 0, li = n - 1; i < li; i++)
			if(mp[i][m / 2] != mp[li][m / 2])
				mult = 2;
	}
	ans *= mult;
	ans %= mod;
	cout << ans << endl;

    return 0;
}