Problem: 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:

<image>

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).

Examples

Input

2 2
cf
cf


Output

6


Input

1 12
codefestival


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
template<class T> using V = vector<T>;
template<class T> using VV = V<V<T>>;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

#ifdef LOCAL
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else 
#define dump(x) true
#endif

constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U> void chmin(T& t, const U& u) { if (t > u) t = u; }
template<class T, class U> void chmax(T& t, const U& u) { if (t < u) t = u; }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

template <unsigned int MOD>
struct ModInt {
	using uint = unsigned int;
	using ull = unsigned long long;
	using M = ModInt;

	uint v;

	ModInt(ll _v = 0) { set_norm(_v % MOD + MOD); }
	M& set_norm(uint _v) { //[0, MOD * 2)->[0, MOD)
		v = (_v < MOD) ? _v : _v - MOD;
		return *this;
	}

	explicit operator bool() const { return v != 0; }
	M operator+(const M& a) const { return M().set_norm(v + a.v); }
	M operator-(const M& a) const { return M().set_norm(v + MOD - a.v); }
	M operator*(const M& a) const { return M().set_norm(ull(v) * a.v % MOD); }
	M operator/(const M& a) const { return *this * a.inv(); }
	M& operator+=(const M& a) { return *this = *this + a; }
	M& operator-=(const M& a) { return *this = *this - a; }
	M& operator*=(const M& a) { return *this = *this * a; }
	M& operator/=(const M& a) { return *this = *this / a; }
	M operator-() const { return M() - *this; }
	M& operator++(int) { return *this = *this + 1; }
	M& operator--(int) { return *this = *this - 1; }

	M pow(ll n) const {
		if (n < 0) return inv().pow(-n);
		M x = *this, res = 1;
		while (n) {
			if (n & 1) res *= x;
			x *= x;
			n >>= 1;
		}
		return res;
	}

	M inv() const {
		ll a = v, b = MOD, p = 1, q = 0, t;
		while (b != 0) {
			t = a / b;
			swap(a -= t * b, b);
			swap(p -= t * q, q);
		}
		return M(p);
	}

	bool operator==(const M& a) const { return v == a.v; }
	bool operator!=(const M& a) const { return v != a.v; }
	friend ostream& operator<<(ostream& os, const M& a) { return os << a.v; }
	static int get_mod() { return MOD; }	
};

using Mint = ModInt<1000000007>;

const int maxv = 100010;

Mint fact[maxv], finv[maxv], inv[maxv], p2[maxv];

void init() {
    fact[0] = 1;
    p2[0] = 1;
    for (int i = 1; i < maxv; ++i) {
        fact[i] = fact[i-1] * i;
        p2[i] = p2[i-1] * 2;
    }

    finv[maxv-1] = fact[maxv-1].inv();
    for (int i = maxv - 2; i >= 0; --i) {
        finv[i] = finv[i+1] * (i+1);
    }

    for (int i = 1; i < maxv; ++i) {
        inv[i] = finv[i] * fact[i-1];
    }
}

Mint comb(int n, int r) {
    if (n < 0 || r < 0 || r > n) return Mint(0);
    return fact[n] * finv[r] * finv[n-r];
}

string s[210];

V<int> g[210];
bool vis[210];

void dfs(int v, int &cnt) {
	vis[v] = 1;
	++cnt;
	for (int to : g[v]) {
		if (!vis[to]) {
			dfs(to, cnt);
		}
	}
}

int main() {
	init();

	int H, W; cin >> H >> W;
	rep(i, H) {
		cin >> s[i];
	}

	int hnum = H / 2, wnum = W / 2;

	Mint coef = 1;

	if (H & 1) {
		string t;
		rep(j, W) t.pb(s[H / 2][j]);
		string rt = t;
		reverse(ALL(rt));
		if (t != rt) {
			coef *= 2;
		}
	}

	if (W & 1) {
		string t;
		rep(i, H) t.pb(s[i][W / 2]);
		string rt = t;
		reverse(ALL(rt));
		if (t != rt) {
			coef *= 2;
		}
	}

	rep(i, hnum) {
		rep(j, wnum) {
			map<char, int> T;
			T[s[i][j]]++;
			T[s[i][W-1-j]]++;
			T[s[H-1-i][W-1-j]]++;
			T[s[H-1-i][j]]++;
			if (T.size() == 4) {
				g[i].pb(j + hnum);
				g[j + hnum].pb(i);
				coef *= 12;
			} else {
				int rem = 4;
				for (auto p : T) {
					coef *= comb(rem, p.se);
					rem -= p.se;
				}
			}
		}
	}

	int cnt = 0;
	rep(i, hnum	+ wnum) {
		if (!vis[i] && g[i].size()) {
			dfs(i, cnt);
			--cnt;
		}
	}
	cout << p2[cnt] * coef << endl;

	return 0;
}