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//
#ifndef __SIZEOF_INT128__
  #define __SIZEOF_INT128__
#endif
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
template <typename T> using oset =  tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define rep(i, p, k) for(int i(p); i < (k); ++i)
#define per(i, p, k) for(int i(p); i > (k); --i)
#define sz(x) (int)(x).size()
#define sc static_cast
typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef __int128_t lll;
//#define int ll
template <typename T = int> using par = std::pair <T, T>;
#define fi first
#define se second
#define test int _number_of_tests(in()); while(_number_of_tests--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb emplace_back
struct Timer {
    string name{""};
    time_point<high_resolution_clock> end, start{high_resolution_clock::now()};
    duration<float, std::milli> dur;
    Timer() = default;
    Timer(string nm): name(nm) {}
    ~Timer() {
        end = high_resolution_clock::now(); dur= end - start;
        cout << "@" << name << "> " << dur.count() << " ms" << '\n';
    }
};
template <typename T = int> inline T in()
{
    static T x;
    std::cin >> x;
    return x;
}
std::string yn(bool b)
{
    if(b) return "YES\n";
    else return "NO\n";
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par);
template <typename T> std::ostream& operator<< (std::ostream& out, const std::vector <T>& wek)
{
    for(const auto& i : wek)out << i << ' ';
    return out;
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par)
{
    out << '{'<<par.first<<", "<<par.second<<"}";
    return out;
}
#define show(x) cerr << #x << " = " << x << '\n';
void solv1(vector <vector <bool>> ar){
    int n(sz(ar)), l(sz(ar[0]));
    int c(1);
    rep(i, 1, n+1) c = lcm(c, i);
    vector <int> pil(l*c);
    rep(i, 0, n)rep(j, 0, l)rep(k, 0, c)pil[j*c+k] += !ar[i][j];
    int p(0), k(l*c);
    while(k > p){
        int S((p+k+1)/2);
        vector <int> pr(n);
        iota(all(pr), 0);
        bool kt = 0;
        vector <vector <int>> mo(n);
        rep(i, 0, n){
            int k(-1), s(0);
            rep(p, 0, l*c){
                while(k - p+1 < S){
                    ++k;
                    if(k == l*c)goto end;
                    s += ar[i][k/c];
                }
                if(!s)mo[i].pb(p);
                s -= ar[i][p/c];
            }
            end:;
        }
        do{
            vector <int> v(pil);
            set <int> zl;
            rep(i, 0, l*c)if(v[i] < 2)zl.insert(i);
            for(auto i: pr){
                for(auto j: mo[i]){
                    auto ptr(zl.lower_bound(j));
                    if(ptr == zl.end() || *ptr >= j+S){
                        rep(k, j, j+S) if(--v[k] < 2)zl.insert(k);
                        goto ok;
                    }
                }
                goto nie;
                ok:;
            }
            kt = 1;
            break;
            nie:;
        }while(next_permutation(all(pr)));
        if(kt)p = S;
        else k = S-1;
    }
    int g(gcd(p,c));
    p /= g;
    c /= g;
    cout << p << '/' << c << '\n';
}
void solv2(vector <vector <bool>> ar){
    int n(sz(ar)), l(sz(ar[0]));
    vector <int> pr{0,0};
    int d(0);
    rep(i, 0, l+1){
        if(i == l || ar[0][i] || ar[1][i]){
            if(d)pr.pb(d);
            d = 0;
        }
        else ++d;
    }
    sort(rall(pr));
    if(pr[0] > pr[1]*2){
        if(pr[0] % 2)cout << pr[0] << '/' << 2 << '\n';
        else cout << pr[0]/2 << '/' << 1 << '\n';
    }
    else cout << pr[1] << '/' << 1 << '\n';
}
std::int32_t main()
{
    std::cout.tie(nullptr); //for luck
    std::cin.tie(nullptr); std::ios_base::sync_with_stdio(0);
    int n(in()), l(in());
    vector <vector <bool>> ar(n, vector <bool>(l));
    rep(i, 0, n)rep(j, 0, l)ar[i][j] = in<char>() == 'X';
    rep(j, 0, l){
        bool b(1);
        rep(i, 0, n)b &= ar[i][j];
        if(b){
            cout << "-1\n";
            return 0;
        }
    }
    if(n == 1)cout << "0/1\n";
    else if(n == 2)solv2(ar);
    else solv1(ar);
    return 0;
}