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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx2")
// 293206GT
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <unordered_set>
#include <stack>
#include <string>
#include<algorithm>
#include <random>
#include <bitset>
#include <chrono>

typedef long long ll;
#define int ll
using namespace std;
#define put(a) cout << (a) << '\n'
#define sqr(x) (x) * (x)
typedef pair<int, int> pii;
typedef long double ld;
typedef pair<ld, ld> pld;
typedef unsigned long long ull;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
#define rep(x, y, a) for (int x = (y); x < (int)(a); ++x)
#define asd(n) rep(i, 0, n)
#define jkl(n) rep(j, 0, n)
#define all(a) a.begin(), a.end()
#define chkmax(a, b) a = max(a, (b))
#define chkmin(a, b) a = min(a, (b))
#define prev asasdasd
#define next aasdasda
#define left asdasdasdasd
#define right asnabsdkasdl
#define rank asdasdknlasd
#define move asdasdas

namespace IO {

    template<class A, class B>
    ostream &operator<<(ostream &out, vector<pair<A, B>> a);

    template<class A>
    ostream &operator<<(ostream &out, vector<A> a);

    template<class A, class B>
    ostream &operator<<(ostream &out, pair<A, B> a) {
        out << a.first << " " << a.second;
        return out;
    }

    template<class A, class B>
    ostream &operator<<(ostream &out, vector<pair<A, B>> a) {
        for (pair<A, B> x: a)
            out << x.first << " " << x.second << '\n';
        return out;
    }

    template<class A>
    ostream &operator<<(ostream &out, vector<A> a) {
        for (A x: a) out << x << ' ';
        return out;
    }

    template<class A, class B>
    istream &operator>>(istream &in, pair<A, B> &a) {
        in >> a.first >> a.second;
        return in;
    }

    template<class A>
    istream &operator>>(istream &in, vector<A> &a) {
        for (A &x: a) in >> x;
        return in;
    }
}
using namespace IO;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void solve() {
    int n, m;
    cin >> n >> m;
    vector<string> a(n);
    cin >> a;
    vpii pl;
    asd(n) jkl(m) {
            if (a[i][j] == 'o') {
                pl.push_back({i, j});
            }
        }
    vvi u(n, vi(m));
    sort(all(pl), [&](pii a, pii b) { return a.first + a.second > b.first + b.second; });
    int ans = 0;
    for (auto[x, y]: pl) {
        vvi d(n, vi(m));
        vvi p(n, vi(m, -1));
        for (int i = n - 1; i >= 0; --i)
            for (int j = m - 1; j >= 0; --j) {
                if (i < n - 1 && !u[i + 1][j] && a[i + 1][j] != '#') {
                    if (d[i][j] < d[i + 1][j] + 1) {
                        d[i][j] = d[i + 1][j] + 1;
                        p[i][j] = 0;
                    }
                }
                if (j < m - 1 && !u[i][j + 1] && a[i][j + 1] != '#') {
                    if (d[i][j] < d[i][j + 1] + 1) {
                        d[i][j] = d[i][j + 1] + 1;
                        p[i][j] = 1;
                    }
                }
            }
        ans += d[x][y];
        while (p[x][y] != -1) {
            if (p[x][y] == 0) ++x;
            else ++y;
        }
        u[x][y] =1;
    }
    put(ans);
}


signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(30);
//    freopen("input.txt", "r", stdin);
    int q = 1;
//    cin >> q;

    while (q--) solve();
}