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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define rep(i,a,n) for(int i=(a);i<=(n);++i)
#define dep(i,a,n) for(int i=(a);i>=(n);--i)
#define eps 1e-8
#define pi 3.1415926535897
#define sqr(x) ((x)*(x))
#define MAX(a,b) a=max(a,b)
#define MIN(a,b) a=min(a,b)
#define SZ(x) ((int)(x).size())
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define CLR(a) memset(a,0,sizeof(a))
#define POSIN(x,y) (1<=(x)&&(x)<=n&&1<=(y)&&(y)<=m)
#define all(x) (x).begin(),(x).end()
#define COUT(S,x) cout<<fixed<<setprecision(x)<<S<<endl
#define buli(x) (__builtin_popcountll(x))
#define bur0(x) (__builtin_ctzll(x))
#define bul2(x) (63-__builtin_clzll(x))
#define pw(x) ((ll(1))<<(x))
#define upmo(a,b) (((a)=((a)+(b))%mo)<0?(a)+=mo:(a))
#define mmo(a,b) (((a)=1ll*(a)*(b)%mo)<0?(a)+=mo:(a))
#define y0 y0z
#define y1 y1z
#define yn ynz
#define j0 j0z
#define j1 j1z
#define jn jnz
#define tm tmz
#ifdef LOCAL
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...) 
#endif
typedef long long ll;
typedef double lf;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<lf,lf> pff;
typedef complex<double> CD;
const int inf=0x3f3f3f3f;
const int mo=1000000007;
inline void gn(long long&x){
	int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
	while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
inline void gn(double&x){double t;scanf("%lf",&t);x=t;}
inline void gn(long double&x){double t;scanf("%lf",&t);x=t;}
template<class T1,class T2>inline void gn(T1&r,T2&s){gn(r),gn(s);}
template<class T1,class T2,class T3>inline void gn(T1&r,T2&s,T3&t){gn(r),gn(s),gn(t);}
template<class T1,class T2,class T3,class T4>inline void gn(T1&r,T2&s,T3&t,T4&u){gn(r),gn(s),gn(t),gn(u);}
inline void gs(char *s){scanf("%s",s);}
inline void gc(char &c){while((c=getchar())>126 || c<33);}
inline void pc(char c){putchar(c);}
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll powmod(ll a,ll b,ll mo) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//*******************************************

const int N=2222,M=111111;
int l,m,n,t,C;
char ch;
int a[N][N],dp[N][N];
set<int>S;
vector<pii>V;
int main(){
#ifdef LOCAL
	freopen("F.in","r",stdin);//freopen("F.out","w",stdout);
#endif
	scanf("%d%d",&m,&n);
	rep(i,1,m)rep(j,1,n)gc(ch),a[i][j]=ch=='#';
	dep(i,m-1,1)rep(j,2,n)if((a[i][j]^a[i][j-1])==(a[i+1][j]^a[i+1][j-1]))dp[i][j]=dp[i+1][j]+1;
	int ans=0;
	rep(i,1,m-1){
		S.clear();
		S.insert(1);S.insert(n+1);
		rep(j,2,n)V.pb(mp(dp[i][j],j));
		sort(V.begin(),V.end());
		for(auto i:V){
			auto p=S.lower_bound(i.Y),q=p;
			p--;
			S.insert(i.Y);
			MAX(ans,(*q-*p)*(i.X+1));
		}
	}
	printf("%d\n",ans);
	return 0;
}