Problem:
Score : 600 points

Problem Statement

There is a board with N rows and M columns.
The information of this board is represented by N strings S_1,S_2,\ldots,S_N.
Specifically, the state of the square at the i-th row from the top and the j-th column from the left is represented as follows:

S_{i,j}=. : the square is empty.

S_{i,j}=# : an obstacle is placed on the square.

S_{i,j}=o : a piece is placed on the square.

Yosupo repeats the following operation:

Choose a piece and move it to its right adjecent square or its down adjacent square.
Moving a piece to squares with another piece or an obstacle is prohibited.
Moving a piece out of the board is also prohibited.

Yosupo wants to perform the operation as many times as possible.
Find the maximum possible number of operations.

Constraints

1 \leq N \leq 50

1 \leq M \leq 50

S_i is a string of length M consisting of ., # and o.

1 \leq ( the number of pieces )\leq 100.
In other words, the number of pairs (i, j) that satisfy S_{i,j}=o is between 1 and 100, both inclusive.

Input

Input is given from Standard Input in the following format:

N M
S_1
S_2
\vdots
S_N

Output

Print the maximum possible number of operations in a line.

Sample Input 1

3 3
o..
...
o.#

Sample Output 1

4

Yosupo can perform operations 4 times as follows:

o.. .o. ..o ... ...
... -> ... -> ... -> ..o -> ..o
o.# o.# o.# o.# .o#

Sample Input 2

9 10
.#....o#..
.#..#..##o
.....#o.##
.###.#o..o
#.#...##.#
..#..#.###
#o.....#..
....###..o
o.......o#

Sample Output 2

24

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
 
#define rep(i,n) for (int i=0;i<(int)(n);++i)
#define rep1(i,n) for (int i=1;i<=(int)(n);++i)
#define range(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define pb push_back
#define F first
#define S second
 
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;
/* 
计算u^v % mod
*/
ll power(ll u,ll v,ll mod){
    ll ret=1,t=u%mod;
    while (v){
        if (v&1) ret=t*ret%mod;
        t=t*t%mod, v>>=1;
    }
    return ret%mod;
}
 
/*
计算gcd(a,b)并找寻ax+by=gcd(a,b)的一组解
ret=exgcd(a,b,x,y); 
*/
ll exgcd(ll a,ll b,ll &x,ll &y){
    if (b){
        ll d=exgcd(b,a%b,y,x);
        return y-=a/b*x, d;
    }
    return x=1,y=0,a;
}
 
/*
计算x=a(mod m), x=b(mod n)的一组解，若无解返回-1
ret=crt(a,m,b,n);
*/
ll crt(ll a,ll m,ll b,ll n){
    if (n>m) swap(a,b), swap(n,m);
    ll x,y,g=exgcd(m,n,x,y);
    if ((b-a)%g) return -1;
    x=(b-a)%n*x%n/g*m+a;
    return x<0?x+m*n/g:x;
}
 
/*
生成<=n的质数
gen_prime<5000>(ret);
*/
 
void gen_prime(int n,vi &ret){
    vector<bool> p;
    p.resize(n+1,1);
    p[0]=p[1]=0;
    for (int i=4;i<=n;i+=2) p[i]=0;
    for (int i=3;i*i<=n;i+=2){
        if (p[i]){
            for (int j=i*i;j<=n;j+=2*i){
                p[j]=0;
            }
        }
    }
    for (int i=2;i<=n;++i){
        if (p[i]) ret.pb(i);
    }
}
 
/*
生成<=n的phi
gen_phi<5000>(ret);
*/
void gen_phi(int n,vi &phi){
    phi.resize(n+1);
    for (int i=0;i<=n;++i) phi[i]=i&1?i:i/2;
    for (int i=3;i<=n;i+=2){
        if (phi[i]==i){
            for (int j=i;j<=n;j+=i) phi[j]-=phi[j]/i;
        }
    }
}
 
/*
支持更高的快速幂操作
expower.mod_pow(a,b,mod);
*/
 
struct expower{
    ull mod_mul(ull a,ull b,ull M) {
        ll ret=a*b-M*ull(ld(a)* ld(b)/ld(M));
        return ret+M*(ret<0)-M*(ret>=(ll)M);
    }
    ull mod_pow(ull b,ull e,ull mod) {
        ull ans=1;
        for (; e; b=mod_mul(b,b,mod), e/=2)
            if (e&1) ans=mod_mul(ans,b,mod);
        return ans%mod;
    }
}expower;
 
/*
Miller-Rubin素数判别
is_prime(n);
*/
bool is_prime(ull n){ 
    if (n<2||n%6%4!=1) return n-2<2;
    ull A[]={2,325,9375,28178,450775,9780504,1795265022},
    s=__builtin_ctzll(n-1), d = n >> s;
    for (auto a:A){
        ull p=expower.mod_pow(a,d,n),i=s;
        while (p!=1&&p!=n-1&&a%n&&i--)
            p=expower.mod_mul(p,p,n);
        if (p!=n-1&&i!=s) return 0;
    }
    return 1;
}
 
/*
素因数分解
ret=factorization(n);
*/
ull pollard(ull n){
    auto f=[n](ull x){return (expower.mod_mul(x,x,n)+1)%n;};
    if (!(n&1)) return 2;
    for (ull i=2;;i++){
        ull x=i,y=f(x),p;
        while ((p=__gcd(n+y-x,n))==1)
            x=f(x),y=f(f(y));
        if (p!=n) return p;
    }
}
 
vector<ull> factorization(ull n){
    if (n==1) return {};
    if (is_prime(n)) return {n};
    ull x=pollard(n);
    auto l=factorization(x),r=factorization(n/x);
    l.insert(l.end(),range(r));
    return l;
}
 
struct DSU{
    //insert info.
    vector<int> p;
    DSU(int n){
        p.resize(n+1);
        for (int i=1;i<=n;++i) p[i]=i;
    }
    int find(int u){
        if (p[u]==u) return u;
        return p[u]=find(find(p[u]));
    } 
    bool unite(int u,int v){
        int pu=find(u), pv=find(v);
        if (pu!=pv){
            p[pu]=pv;
        }
        return pu!=pv;
    }
};
 
/* (u,v,w) */
template<class T> vector<tuple<int,int,T> > MST(int n,vector<tuple<int,int,T> >g){
    vector<tuple<int,int,T> > ret;
    ret.clear();
    DSU dsu(n);
    auto cmp=[&](tuple<int,int,T> x,tuple<int,int,T> y){
        int ux,uy,vx,vy;
        T wx,wy;
        tie(ux,vx,wx)=x, tie(uy,vy,wy)=y;
        return wx<wy;
    };
    sort(range(g),cmp);
    for (auto c:g){
        int u,v;
        T w;
        tie(u,v,w)=c;
        if(dsu.unite(u,v)) ret.push_back(c);
    }
    return ret;
}
 
template<int n,class T>
struct SegmentTree{
    typedef struct{
        T lz,mx,mn,sum;
    }Node;
    vector<Node> tree;
    SegmentTree(){
        tree.resize((n+7)<<2);
    }  
    void pushup(int num){
        tree[num].mx=max(tree[num<<1].mx,tree[num<<1|1].mx);
        tree[num].mn=min(tree[num<<1].mn,tree[num<<1|1].mn);
        tree[num].sum=tree[num<<1].sum+tree[num<<1|1].sum;
    }
    void pushdown(int num,int l,int r,int md){
        if (tree[num].lz){
            tree[num<<1].mx+=tree[num].lz;
            tree[num<<1|1].mx+=tree[num].lz;
            tree[num<<1].mn+=tree[num].lz;
            tree[num<<1|1].mn+=tree[num].lz;
            tree[num<<1].sum+=tree[num].lz*(md-l+1);
            tree[num<<1|1].sum+=tree[num].lz*(r-md);
            tree[num<<1].lz+=tree[num].lz;
            tree[num<<1|1].lz+=tree[num].lz;
            tree[num].lz=0;
        }
    }
    void update1(int num,int l,int r,int L,int R,T val){
        if (L<=l&&r<=R){
            tree[num].mx+=val;
            tree[num].mn+=val;
            tree[num].sum+=val*(r-l+1);
            tree[num].lz+=val;
            return;
        }
        int md=(l+r)>>1;
        pushdown(num,l,r,md);
        if (md>=L) update1(num<<1,l,md,L,R,val);
        if (md<R) update1(num<<1|1,md+1,r,L,R,val);
        pushup(num);
    }
    tuple<T,T,T> query1(int num,int l,int r,int L,int R){
        if (L<=l&&r<=R){
            return {tree[num].mn,tree[num].mx,tree[num].sum};
        }
        int md=(l+r)>>1;
        pushdown(num,l,r,md);
        if (md>=R){
            return query1(num<<1,l,md,L,R);
        }
        else{
            if (md<L){
                return query1(num<<1|1,md+1,r,L,R);
            }
            else{
                tuple<T,T,T> ret1=query1(num<<1,l,md,L,R), ret2=query1(num<<1|1,md+1,r,L,R);
                T a,b,c,d,e,f;
                tie(a,b,c)=ret1, tie(d,e,f)=ret2;
                return {min(a,d),max(b,e),c+f};
            }
        } 
    }
    void update(int L,int R,int w){
        update1(1,1,n,L,R,w);
    }
    tuple<T,T,T> query(int L,int R){
        return query1(1,1,n,L,R);
    }
    void debug1(int num,int l,int r){
        if (l==r){
            cout<<tree[num].mn<<" "; return;
        }
        int md=(l+r)>>1;
        pushdown(num,l,r,md);
        debug1(num<<1,l,md);
         debug1(num<<1|1,md+1,r);        
    }
    void debug(){
        debug1(1,1,n);
        cout<<endl;
    }
};
 
struct XORBasis{
    int n;
    vector<ull> p;
    vector<ull> ans;
    XORBasis(int k):n(k){
        p.resize(k+1);
    }
    void rebuilt(){
        ans.clear();
        for (int i=0;i<=n;++i){
            if (p[i]) ans.push_back(p[i]);
        }
    }
    void insert(ull u){
        for (int i=n;i>=0;--i){
            if (u&(1ull<<i)){
                if (p[i]) u^=p[i];
                else{
                    p[i]=u;
                    for (int j=i-1;j>=0;--j){
                        if (p[i]&(1ull<<j)) p[i]^=p[j];
                    }
                    for (int j=i+1;j<=n;++j){
                        if (p[j]&(1ull<<i)) p[j]^=p[i];
                    }
                    break;
                }
            }
        }
        rebuilt();
    }
    bool check(ull u){
        for (int i=n;i>-1;--i){
            if (u&(1ull<<i)) u^=p[i];
        }
        return (u==0);
    }
    ull kth(ull t){ // the k-th smallest, excluding 0.
        if (t>=(1ull<<sz(ans))) return -1;
        ull ret=0;
        for (int i=0;i<sz(ans);++i){
            if (t&(1ull<<i)) ret^=ans[i];
        }
        return ret;
    }
    ull total(){
        return (1ull<<sz(ans));
    }
};
 
 
vi pi(const string& s) {
	vi p(sz(s));
	for (int i=1;i<=sz(s);++i) {
		int g = p[i-1];
		while (g && s[i] != s[g]) g = p[g-1];
		p[i] = g + (s[i] == s[g]);
	}
	return p;
}
 
vi match(const string& s, const string& pat) {
	vi p = pi(pat + '\0' + s), res;
	for(int i=sz(p)-sz(s);i<=sz(p);++i)
		if (p[i] == sz(pat)) res.push_back(i - 2 * sz(pat));
	return res;
}
 
 
struct AhoCorasick {
	enum {alpha = 26, first = 'a'};
	struct Node {
		// (nmatches is optional)
		int back, next[alpha], start = -1, end = -1;
        ll nmatches = 0;
		Node(int v) { memset(next, v, sizeof(next)); }
	};
	vector<Node> N;
	vector<int> backp;
	void insert(string& s, int j,ll val) {
		assert(!s.empty());
		int n = 0;
		for(auto c:s) {
			int& m = N[n].next[c - first];
			if (m == -1) { n = m = sz(N); N.emplace_back(-1); }
			else n = m;
		}
		if (N[n].end == -1) N[n].start = j;
		backp.push_back(N[n].end);
		N[n].end = j;
		N[n].nmatches+=val;
	}
	AhoCorasick(vector<string>& pat,vi val) {
		N.emplace_back(-1);
		for (int i=0;i<sz(pat);++i) insert(pat[i], i, val[i]);
		N[0].back = sz(N);
		N.emplace_back(0);
 
		queue<int> q;
		for (q.push(0); !q.empty(); q.pop()) {
			int n = q.front(), prev = N[n].back;
			for (int i=0;i<alpha;++i) {
				int &ed = N[n].next[i], y = N[prev].next[i];
				if (ed == -1) ed = y;
				else {
					N[ed].back = y;
					(N[ed].end == -1 ? N[ed].end : backp[N[ed].start])
						= N[y].end;
					N[ed].nmatches += N[y].nmatches;
					q.push(ed);
				}
			}
		}
	}
	vi find(string& word,ll& ans) {
		int n = 0;
		vi res;
		for(auto c:word) {
			n = N[n].next[c - first];
			res.push_back(N[n].end);
			ans += N[n].nmatches;
		}
		return res;
	}
	vector<vi> findAll(vector<string>& pat, string word) {
		ll ans=0;
        vi r = find(word,ans);
		vector<vi> res(sz(word));
		for (int i=0;i<sz(word);++i) {
			int ind = r[i];
			while (ind != -1) {
				res[i - sz(pat[ind]) + 1].push_back(ind);
				ind = backp[ind];
			}
		}
		return res;
	}
};
 
struct HLD{
    int n,cnt=0;
    vector<vi> G;
    vi par,dep,siz,son,top,dfn,rk; 
    //父亲节点，深度，子树大小，重子节点，重链头部，dfs序，dfs序对应的节点编号
    HLD(vector<vi> g,int root=1):n(sz(g)),G(g),par(n+7),dep(n+7),siz(n+7,1),son(n+7,-1),top(n+7),dfn(n+7),rk(n+7){
        dep[0]=-1, dfs1(root,0), dfs2(root,0,root);
    }
    void dfs1(int u,int p){
        for (auto c:G[u]){
            if (c==p) continue;
            dep[c]=dep[u]+1, par[c]=u;
            dfs1(c,u);
            siz[u]+=siz[c];
            if (son[u]==-1||siz[son[u]]<siz[c]) son[u]=c;
        }
    }
    void dfs2(int u,int p,int tp){
        top[u]=tp;
        dfn[u]=++cnt;
        rk[cnt]=u;
        if (son[u]==-1) return;
        dfs2(son[u],u,tp);
        for (auto c:G[u]){
            if (c==p||c==son[u]) continue;
            dfs2(c,u,c);
        }
    }
    int lca(int u,int v){
        while (top[u]!=top[v]){
            if (dep[top[u]]>dep[top[v]]){
                u=par[top[u]];
            }
            else{
                v=par[top[v]];
            }
        }
        return dep[u]>dep[v]?v:u;
    }
    vector<pii> find_seg(int u,int v){
        vector<pii> ft,bk,res;
        ft.clear(), bk.clear(), res.clear();
//        cout<<top[u]<<" "<<top[v]<<endl;
        while (top[u]!=top[v]){
            if (dep[top[u]]>dep[top[v]]){
                ft.pb({u,top[u]});
                u=par[top[u]];
            }
            else{
                bk.pb({top[v],v});
                v=par[top[v]];
            }
        }
        ft.pb({u,v});
        for (int i=0;i<sz(ft);++i) res.pb(ft[i]);
        for (int i=sz(bk);i>0;--i) res.pb(bk[i-1]);
        return res;
    }
};
 
struct BIT{
    vi C; int n;
    BIT(int n):n(n){C.resize(n+7,0);}
    void update(int u,int w){
        for (;u<=n;u+=u&(-u)) C[u]+=w;
    }
    int query(int u){
        int res=0;
        for (;u>0;u-=u&(-u)) res+=C[u];
        return res; 
    }
};

vi topoSort(const vector<vi>& gr) {
	vi indeg(sz(gr)), ret;
	for (auto& li : gr) for (int x : li) indeg[x]++;
	queue<int> q; // use priority queue for lexic. smallest ans.
	rep(i,sz(gr)) if (indeg[i] == 0) q.push(-i);
	while (!q.empty()) {
		int i = -q.front(); // top() for priority queue
		ret.push_back(i);
		q.pop();
		for (int x : gr[i])
			if (--indeg[x] == 0) q.push(-x);
	}
	return ret;
}

vi path;
namespace Eulerian {
    const int N = 3, M = 3;
    int t, n, m, tot, cc, hd[N], nxt[M], to[M], bh[M], vis[M], in[N], out[N], ans[M];
    void add(int x, int y, int id) {
        nxt[++tot] = hd[x];
        hd[x] = tot;
        to[tot] = y;
        bh[tot] = id;
        in[y]++;
        out[x]++;
    }
    void dfs(int x) {
        for (int i = hd[x]; i; i = hd[x]) {
            while (i && vis[abs(bh[i])]) i = nxt[i];
            hd[x] = i;
            if (!i) break;
            vis[abs(bh[i])] = 1;
            dfs(to[i]), ans[++cc] = bh[i];
        }
    }
 
    void init() {
        memset(hd, 0, sizeof(hd));
        memset(nxt, 0, sizeof(nxt));
        memset(to, 0, sizeof(to));
        memset(bh, 0, sizeof(bh));
        memset(vis, 0, sizeof(vis));
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        memset(ans, 0, sizeof(ans));
        t = 1;
        n = 0;
        m = 0;
        tot = 0;
        cc = 0;
    }
    bool solve() {
        for (int i = 1; i <= n; i++)
            if ((t == 1 && in[i] & 1)) {
                return false;
            }
        dfs(to[1]);
        if (cc < m) return false;
        else {
            path.clear();
            for (int i = cc; i >= 1; i--) path.push_back(ans[i]);
            return true;
        }
    }
} // namespace Eulerian

namespace FWT{
    //mod=-1 (no mod)
    const int mod=-1;

    inline int modadd(int a,int b){
        a+=b;
        if (mod==-1) return a;
        a%=mod, a+=mod, a%=mod;
        return a;
    }

    vi OR(vi &a,int mode=1){
        vi b(a);
        for (int i=0;(1<<i)<sz(a);++i){
            for (int j=0;j<sz(a);++j){
                if (j>>i&1) b[j]=modadd(b[j],b[j^(1<<i)]*mode); 
            }
        }
        return b;
    }
    vi AND(vi &a,int mode=1){
        vi b(a);
        for (int i=0;(1<<i)<sz(a);++i){
            for (int j=0;j<sz(a);++j){
                if (j>>i&1) b[j^(1<<i)]=modadd(b[j^(1<<i)],b[j]*mode);
            }
        }
        return b;
    }
    vi XOR(vi &a,int mode=1){
        vi b(a);
        for (int i=0;(1<<i)<sz(a);++i){
            for (int j=0;j<sz(a);++j){
                if (j>>i&1){
                    b[j^(1<<i)]=modadd(b[j^(1<<i)],b[j]);
                    b[j]=modadd(b[j^(1<<i)],-2*b[j]);
                    if (mode<0) {
                        if (mod>0) b[j]=b[j]*power(2,mod-2,mod)%mod, b[j^(1<<i)]=b[j^(1<<i)]*power(2,mod-2,mod)%mod;
                        else b[j]/=2, b[j^(1<<i)]/=2;
                    }
                }
            }
        }
        return b;
    }
}

/* atcoder mincost-maxflow */

template <class Cap, class Cost> struct mcf_graph {
  public:
    mcf_graph() {}
    mcf_graph(int n) : _n(n), g(n) {}

    int add_edge(int from, int to, Cap cap, Cost cost) {
        assert(0 <= from && from < _n);
        assert(0 <= to && to < _n);
        int m = (int)(pos.size());
        pos.push_back({from, (int)(g[from].size())});
        g[from].push_back(_edge{to, (int)(g[to].size()), cap, cost});
        g[to].push_back(_edge{from, (int)(g[from].size()) - 1, 0, -cost});
        return m;
    }

    struct edge {
        int from, to;
        Cap cap, flow;
        Cost cost;
    };

    edge get_edge(int i) {
        int m = (int)(pos.size());
        assert(0 <= i && i < m);
        auto _e = g[pos[i].first][pos[i].second];
        auto _re = g[_e.to][_e.rev];
        return edge{
            pos[i].first, _e.to, _e.cap + _re.cap, _re.cap, _e.cost,
        };
    }
    std::vector<edge> edges() {
        int m = (int)(pos.size());
        std::vector<edge> result(m);
        for (int i = 0; i < m; i++) {
            result[i] = get_edge(i);
        }
        return result;
    }

    std::pair<Cap, Cost> flow(int s, int t) {
        return flow(s, t, std::numeric_limits<Cap>::max());
    }
    std::pair<Cap, Cost> flow(int s, int t, Cap flow_limit) {
        return slope(s, t, flow_limit).back();
    }
    std::vector<std::pair<Cap, Cost>> slope(int s, int t) {
        return slope(s, t, std::numeric_limits<Cap>::max());
    }
    std::vector<std::pair<Cap, Cost>> slope(int s, int t, Cap flow_limit) {
        assert(0 <= s && s < _n);
        assert(0 <= t && t < _n);
        assert(s != t);
        // variants (C = maxcost):
        // -(n-1)C <= dual[s] <= dual[i] <= dual[t] = 0
        // reduced cost (= e.cost + dual[e.from] - dual[e.to]) >= 0 for all edge
        std::vector<Cost> dual(_n, 0), dist(_n);
        std::vector<int> pv(_n), pe(_n);
        std::vector<bool> vis(_n);
        auto dual_ref = [&]() {
            std::fill(dist.begin(), dist.end(),
                      std::numeric_limits<Cost>::max());
            std::fill(pv.begin(), pv.end(), -1);
            std::fill(pe.begin(), pe.end(), -1);
            std::fill(vis.begin(), vis.end(), false);
            struct Q {
                Cost key;
                int to;
                bool operator<(Q r) const { return key > r.key; }
            };
            std::priority_queue<Q> que;
            dist[s] = 0;
            que.push(Q{0, s});
            while (!que.empty()) {
                int v = que.top().to;
                que.pop();
                if (vis[v]) continue;
                vis[v] = true;
                if (v == t) break;
                // dist[v] = shortest(s, v) + dual[s] - dual[v]
                // dist[v] >= 0 (all reduced cost are positive)
                // dist[v] <= (n-1)C
                for (int i = 0; i < (int)(g[v].size()); i++) {
                    auto e = g[v][i];
                    if (vis[e.to] || !e.cap) continue;
                    // |-dual[e.to] + dual[v]| <= (n-1)C
                    // cost <= C - -(n-1)C + 0 = nC
                    Cost cost = e.cost - dual[e.to] + dual[v];
                    if (dist[e.to] - dist[v] > cost) {
                        dist[e.to] = dist[v] + cost;
                        pv[e.to] = v;
                        pe[e.to] = i;
                        que.push(Q{dist[e.to], e.to});
                    }
                }
            }
            if (!vis[t]) {
                return false;
            }

            for (int v = 0; v < _n; v++) {
                if (!vis[v]) continue;
                // dual[v] = dual[v] - dist[t] + dist[v]
                //         = dual[v] - (shortest(s, t) + dual[s] - dual[t]) + (shortest(s, v) + dual[s] - dual[v])
                //         = - shortest(s, t) + dual[t] + shortest(s, v)
                //         = shortest(s, v) - shortest(s, t) >= 0 - (n-1)C
                dual[v] -= dist[t] - dist[v];
            }
            return true;
        };
        Cap flow = 0;
        Cost cost = 0, prev_cost = -1;
        std::vector<std::pair<Cap, Cost>> result;
        result.push_back({flow, cost});
        while (flow < flow_limit) {
            if (!dual_ref()) break;
            Cap c = flow_limit - flow;
            for (int v = t; v != s; v = pv[v]) {
                c = std::min(c, g[pv[v]][pe[v]].cap);
            }
            for (int v = t; v != s; v = pv[v]) {
                auto& e = g[pv[v]][pe[v]];
                e.cap -= c;
                g[v][e.rev].cap += c;
            }
            Cost d = -dual[s];
            flow += c;
            cost += c * d;
            if (prev_cost == d) {
                result.pop_back();
            }
            result.push_back({flow, cost});
            prev_cost = cost;
        }
        return result;
    }

  private:
    int _n;

    struct _edge {
        int to, rev;
        Cap cap;
        Cost cost;
    };

    std::vector<std::pair<int, int>> pos;
    std::vector<std::vector<_edge>> g;
};

const int maxn=200007;
 
int n,m,cnt;
string s[57];
mcf_graph<int,int> dinic(3000);
 
void dfs(int i,int j,int ini,int inj){
    dinic.add_edge(n*m+cnt+1,i*m+j+2,1,n+m-(i-ini+j-inj));
 //   cout<<cnt<<" "<<i<<" "<<j<<" "<<i-ini+j-inj<<endl;
    if (i+1<n&&s[i+1][j]!='#') dfs(i+1,j,ini,inj);
    if (j+1<m&&s[i][j+1]!='#') dfs(i,j+1,ini,inj);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin>>n>>m;
    cnt=0;
    rep(i,n) cin>>s[i];
    rep(i,n) rep(j,m){
        if (s[i][j]=='o') ++cnt, dfs(i,j,i,j);
    }
    for (int i=1;i<=cnt;++i) dinic.add_edge(1,n*m+i+1,1,0);
    rep(i,n) rep(j,m) dinic.add_edge(i*m+j+2,n*m+cnt+7,1,0);
 //   cout<<dinic.maxflow(1,n*m+cnt+7).first<<endl;
    cout<<(n+m)*cnt-(dinic.flow(1,n*m+cnt+7).second);
    return 0;    
}