Problem:
Score : 700 points

Problem Statement

We have a board with an H \times W grid.
Each square in the grid is painted in black or white. The square at the i-th row from the top and j-th column from the left is black if the j-th character in S_i is #, and white if that character is ..

Snuke can perform the following operation on the grid any number of times:

Select a row or column in the grid, and invert the color of all the squares in that row or column (that is, black squares become white and vice versa).

Then, Snuke draws a rectangle along grid lines. Here, all the squares contained in the rectangle must be painted in black.

Find the maximum possible area of Snuke's rectangle when the operation is performed optimally.

Constraints

2 \leq H \leq 2000

2 \leq W \leq 2000

|S_i| = W

S_i consists of # and ..

Input

Input is given from Standard Input in the following format:

H W
S_1
S_2
:
S_H

Output

Print the maximum possible area of Snuke's rectangle.

Sample Input 1

3 3
..#
##.
.#.

Sample Output 1

6

If the first row from the top and the third column from the left are inverted, a 2 \times 3 rectangle can be drawn, as shown below:

Sample Input 2

4 4
....
....
....
....

Sample Output 2

16

Sample Input 3

10 8
##...#.#
##...#.#
..###.#.
#.##.#.#
.#..#.#.
..##.#.#
##.#.#..
...#.#..
###.#.##
###..###

Sample Output 3

27

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#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 __gnu_pbds;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
constexpr char ln = '\n';
constexpr long long MOD = 1000000007;
//constexpr long long MOD = 998244353;
template<class T1, class T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return true;} return false; }
template<class T1, class T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return true;} return false; }
inline int popcount(int x) {return __builtin_popcount(x);}
inline int popcount(long long x) {return __builtin_popcountll(x);}
void print() { cout << "\n"; }
template<class T, class... Args>
void print(const T &x, const Args &... args) {
    cout << x << " ";
    print(args...);
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    auto solve=[](vector<string> S) {
        int N = S.size();
        int M = S[0].size();
        int ret = 0;
        vector<vector<int>> A(N, vector<int>(M-1,1));
        rep(i,N-1) {
            rep(j,M-1) {
                int a = S[i][j]==S[i][j+1];
                int b = S[i+1][j]==S[i+1][j+1];
                if (a^b) continue;
                A[i+1][j] = A[i][j]+1;
            }
        }
        rep(i,N) {
            vector<pii> sta;
            rep(j,M-1) {
                int pos = j;
                if (!sta.empty() and A[i][j] < sta.back().first) {
                    while (!sta.empty() and sta.back().first >= A[i][j]) {
                        auto[h,x] = sta.back(); sta.pop_back();
                        chmax(ret,h*(j-x+1));
                        pos = x;
                    }                
                }
                if (sta.empty() or A[i][j] > sta.back().first) {
                    sta.emplace_back(A[i][j],pos);
                }
            }
            while (!sta.empty()) {
                auto[h,x] = sta.back(); sta.pop_back();
                chmax(ret,h*(M-1-x+1));
            }
        }

        return ret;
    };

    int N,M; cin >> N >> M;
    vector<string> S(N);
    rep(i,N) cin >> S[i];

    int ans = solve(S);
    vector<string> T(M, string(N,' '));
    rep(i,N) rep(j,M) T[j][i] = S[i][j];
    chmax(ans,solve(T));

    cout << ans << ln;
}