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 <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("0"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;

// Mod int
const int mod = 1000000007;
struct mint;
mint ex(mint a, ll tms);
struct mint {
  ll x;
  mint():x(0){}
  mint(ll x):x((x%mod+mod)%mod){}
  // mint(ll x):x(x){}
  mint& fix() { x = (x%mod+mod)%mod; return *this;}
  mint& operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a){ (x*=a.x)%=mod; return *this;}
  mint& operator/=(const mint& a){ (x*=ex(a,mod-2).x)%=mod; return *this;}
  mint operator+(const mint& a)const{ return mint(*this) += a;}
  mint operator-(const mint& a)const{ return mint(*this) -= a;}
  mint operator*(const mint& a)const{ return mint(*this) *= a;}
  mint operator/(const mint& a)const{ return mint(*this) /= a;}
  bool operator<(const mint& a)const{ return x < a.x;}
  bool operator==(const mint& a)const{ return x == a.x;}
};
istream& operator>>(istream&i,mint&a){i>>a.x;return i;}
ostream& operator<<(ostream&o,const mint&a){o<<a.x;return o;}
typedef vector<mint> vm;
mint ex(mint a, ll tms) {
  if(!tms) return 1;
  mint res = ex(a,tms/2);
  res *= res;
  return (tms&1)?res*a:res;
}
struct comb {
  vm f, g;
  comb(){}
  comb(int mx):f(mx+1),g(mx+1) {
    f[0] = 1;
    rrep(i,mx) f[i] = f[i-1]*i;
    g[mx] = ex(f[mx],mod-2);
    for(int i=mx;i>0;i--) g[i-1] = g[i]*i;
  }
  mint c(int a, int b) {
    if (a < b) return 0;
    return f[a]*g[b]*g[a-b];
  }
};
//

// Union find
struct uf {
  vi d;
  uf(){}
  uf(int mx):d(mx,-1){}
  int root(int x) {
    if(d[x] < 0) return x;
    return d[x] = root(d[x]);
  }
  bool unite(int x, int y) {
    x = root(x); y = root(y);
    if(x == y) return false;
    if(d[x] > d[y]) swap(x,y);
    d[x] += d[y]; d[y] = x;
    return true;
  }
  int size(int x) { return -d[root(x)];}
};
//

bool pari(string s) {
  string t = s;
  reverse(rng(t));
  return s == t;
}

int main() {
  df(h); df(w);
  scanf("%d%d",&h,&w);
  vector<string> s(h);
  rep(i,h) cin >> s[i];
  mint ans = 1;
  if (h&1) {
    if (!pari(s[h/2])) ans *= 2;
  }
  if (w&1) {
    string t;
    rep(i,h) t += s[i][w/2];
    if (!pari(t)) ans *= 2;
  }
  uf d(h/2+w/2);
  rep(i,h/2)rep(j,w/2) {
    string t;
    t += s[i][j];
    t += s[i][w-1-j];
    t += s[h-1-i][j];
    t += s[h-1-i][w-1-j];
    sort(rng(t));
    vi a;
    char pre = '$';
    for (char c : t) {
      if (pre != c) a.pb(0);
      a.back()++;
      pre = c;
    }
    if (sz(a) == 4) {
      ans *= 12;
      if (d.unite(i,h/2+j)) ans *= 2;
    } else {
      ans *= 24;
      for (int x : a) rrep(k,x) ans /= k;
    }
  }
  cout << ans << endl;
  return 0;
}