Main idea of the title:

Given a 01 rectangle of n * m n times m n * m, one row or one column can be flipped at a time.
Find the maximum all-1 sub-rectangle after several flips.

Train of thought:

First of all, we need to know a conclusion: if a sub-rectangle can be inverted into a full-1 rectangle, then the number of 1 of each sub-rectangle within it is even.
If there exists an odd number of 1 in a 2*22\times 22*2 sub-rectangle, then any operation is odd.
If all 2 *22\times 22 *2 sub-rectangles have even numbers of 1, we can first make the first row and first column of the rectangle become 1. According to parity, it is not difficult to find that the whole matrix must all become 1 at this time.
So we just need to find a maximum rectangle containing only even number 1, which can be transformed into a classical maximum all-1 sub-matrix problem. With monotone stack maintenance, we can achieve O(n2)O(n^2)O(n2).
Notice that the minimum answer is max (n, m) max (n, m) max (n, m), and finally remember chkmax.

#include<bits/stdc++.h>

#define REP(i,a,b) for(int i=a,i##_end_=b;i<=i##_end_;++i)
#define DREP(i,a,b) for(int i=a,i##_end_=b;i>=i##_end_;--i)
#define MREP(i,x) for(int i=beg[x],v;v=to[i],i;i=las[i])
#define debug(x) cout<<#x<<"="<<x<<endl
#define fi first
#define se second
#define mk make_pair
#define pb push_back
#define y1 asdasd
typedef long long ll;

using namespace std;

void File(){
	freopen("speech.in","r",stdin);
	freopen("speech.out","w",stdout);
}

template<typename T>void read(T &_){
	T __=0,mul=1; char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-')mul=-1;
		ch=getchar();
	}
	while(isdigit(ch))__=(__<<1)+(__<<3)+(ch^'0'),ch=getchar();
	_=__*mul;
}

const int maxn=2000+10;
int n,m,a[maxn][maxn],b[maxn][maxn],ans;
char s[maxn];
stack<int>stk;

int main(){
	File();
	read(n); read(m);
	if(n==1 || m==1)return printf("%d\n",n*m),0;
	REP(i,1,n){
		scanf("%s",s+1);
		REP(j,1,m)a[i][j]=(s[j]=='#');
	}
	REP(i,1,n-1)REP(j,1,m-1){
		int c[2]={0};
		++c[a[i][j]];
		++c[a[i+1][j]];
		++c[a[i][j+1]];
		++c[a[i+1][j+1]];
		b[i][j]=!(c[1]%2);
		if(b[i][j])b[i][j]+=b[i-1][j];
	}
	REP(i,1,n-1){
		int las;
		REP(j,1,m-1){
			las=j;
			while(!stk.empty() && b[i][stk.top()]>=b[i][j]){
				las=stk.top(); stk.pop();
				if(b[i][las])ans=max(ans,(j-las+1)*(b[i][las]+1));
			}
			b[i][las]=b[i][j];
			stk.push(las);
		}
		while(!stk.empty()){
			las=stk.top(); stk.pop();
			if(b[i][las])ans=max(ans,(m-las+1)*(b[i][las]+1));
		}
	}
	printf("%d\n",max(ans,max(n,m)));
	return 0;
}