#include <iostream>
#include <string>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <tuple>
#include <map>
#include<set>
#include<queue>
#include<stack>
#include <unordered_set>
#include<thread>
#include<bits/stdc++.h>

#include <atcoder/all>
#include <cstdio>

#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

//if(a < 0 || h <= a || b < 0 || w <= b)return;
// string abc = "abcdefghijklmnopqrstuvwxyz";
// string abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

using namespace std;
using namespace atcoder;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using mint = modint998244353;
//using mint = modint1000000007;
template<typename T> using pq = priority_queue<T>;//降順?(最大取り出し)
template<typename T> using pqg = priority_queue<T, vector<T>, greater<T>>;//昇順?(最小取り出し)
template<typename T> using vector2 = vector<vector<T>>;
template<typename T> using vector3 = vector<vector<vector<T>>>;
template<typename T> using vector4 = vector<vector<vector<vector<T>>>>;
template<typename T> using vector5 = vector<vector<vector<vector<vector<T>>>>>;
template<typename T> using vector6 = vector<vector<vector<vector<vector<vector<T>>>>>>;
template<typename T> using pairs = pair<T,T>;
#define rep(i, n) for (ll i = 0; i < ll(n); i++)
#define rep1(i,n) for(int i = 1;i <= int(n);i++)
#define repm(i, m, n) for (int i = (m); (i) < int(n);(i)++)
#define repmr(i, m, n) for (int i = (m) - 1; (i) >= int(n);(i)--)
#define rep0(i,n) for(int i = n - 1;i >= 0;i--)
#define rep01(i,n) for(int i = n;i >= 1;i--)




// ユークリッドの互除法による最大公約数算出
ll GCD(ll a,ll b){
    if(b == 0)return a;
    return GCD(b, a % b);
}
//拡張ユークリッドの互除法による(ax + by = GCD(a,b))を満たすx,yの算出
pair<long long, long long> extgcd(long long a, long long b) {
  if (b == 0) return make_pair(1, 0);
  long long x, y;
  tie(y, x) = extgcd(b, a % b);
  y -= a / b * x;
  return make_pair(x, y);
}
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<int> nu;
    
    UnionFind(int N) : par(N) , nu(N){ //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
        for(int i = 0;i < N;i++)nu[i] = 1;
    }
    
    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }
    
    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        nu[rx] += nu[ry];
        nu[ry] = nu[rx];
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }
    
    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }

    int num(int x){//根がxである頂点の数を返す
        return nu[x];
    }
};
ll n;


//座標圧縮
vector<ll> Ccomp(vector<ll> a){
    vector<ll> b = a;
    sort(b.begin(),b.end());
    b.erase(unique(b.begin(),b.end()),b.end());//ダブり消去
    vector<ll> rtn;
    rep(j,a.size()){
        ll pb = lower_bound(b.begin(),b.end(),a[j]) - b.begin();
        rtn.push_back(pb);
    }
    return rtn;
}
/// ここから////////////////////////////////////////////




using F = ll;
using S = ll;
string s;

ll modPow(ll a, ll n, ll mod) { if(mod==1) return 0;ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1; } return ret; }
void cincout(){
  ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
  cout<< fixed << setprecision(15);
}
//seg,遅延segの設定-----ここから
S op(S a,S b){return max(a,b);}//何を求めるか(最大値とか)

S e(){return 0;}//モノイド(初期値)

S mapping (F a,S b){return a + b;}//遅延処理

F composition (F a,F b){return a + b;}//遅延中の枝にさらに処理

F id(){return 0;}//遅延のモノイド

vector<int> Op(vector<int> a,vector<int> b){a.insert(a.end(),b.begin(),b.end()); return a;}

vector<int> E(){return vector<int> (0);}


ll INF = ll(2e9);

map<vector<int>,int> maps;



int main() {
    cincout();
    ll h,w;
    cin >> h >> w;
    vector2<int> a(h,vector<int>(w));
    vector2<int> b(h,vector<int>(w));
    vector2<int> c(h,vector<int>(w));
    //vector2<int> b(h,vector<int>(w));
    rep(j,h){
        string s;
        cin >> s;
        rep(i,w){
            if(s.substr(i,1) == "#")a[j][i] = 1;
            else a[j][i] = 0;
        }
    }
    rep(j,h){
        cin >> s;
        rep(i,w){
            if(s.substr(i,1) == "#")b[j][i] = 0;
            else b[j][i] = 1;
        }
    }
    rep(j,h){
        rep(i,w){
            c[j][i] = a[h-j-1][w - i - 1];
        }
    }

    bool st = a == b,sc = c == b;


    double ans = 0;
    double now = 1;
    double noww = 1;
    if(!st && !sc){
        cout << -1 << endl;
        return 0;
    }
    rep(j,1e5){
        //cout << ans << " " << now << endl;
        if(j % 2 == 0){
            if(st){
                ans += (j + 1) * noww * (1 - now);
                noww *= now;
            }
        }
        else{
            if(sc){
                ans += (j + 1) * noww *  (1 - now);
                noww *= now;
            }
        }
        now /= 2;
    }
    cout << ans << endl;
    return 0;
}