Problem:
Score : 1000 points

Problem Statement

Problem F and F2 are the same problem, but with different constraints and time limits.

We have a board divided into N horizontal rows and N vertical columns of square cells.
The cell at the i-th row from the top and the j-th column from the left is called Cell (i,j).
Each cell is either empty or occupied by an obstacle.
Also, each empty cell has a digit written on it.
If A_{i,j}= 1, 2, ..., or 9, Cell (i,j) is empty and the digit A_{i,j} is written on it.
If A_{i,j}= #, Cell (i,j) is occupied by an obstacle.

Cell Y is reachable from cell X when the following conditions are all met:

Cells X and Y are different.

Cells X and Y are both empty.

One can reach from Cell X to Cell Y by repeatedly moving right or down to an adjacent empty cell.

Consider all pairs of cells (X,Y) such that cell Y is reachable from cell X.
Find the sum of the products of the digits written on cell X and cell Y for all of those pairs.

Constraints

1 \leq N \leq 500

A_{i,j} is one of the following characters: 1, 2, ... 9 and #.

Input

Input is given from Standard Input in the following format:

N
A_{1,1}A_{1,2}...A_{1,N}
A_{2,1}A_{2,2}...A_{2,N}
:
A_{N,1}A_{N,2}...A_{N,N}

Output

Print the sum of the products of the digits written on cell X and cell Y for all pairs (X,Y) such that cell Y is reachable from cell X.

Sample Input 1

2
11
11

Sample Output 1

5

There are five pairs of cells (X,Y) such that cell Y is reachable from cell X, as follows:

X=(1,1), Y=(1,2)

X=(1,1), Y=(2,1)

X=(1,1), Y=(2,2)

X=(1,2), Y=(2,2)

X=(2,1), Y=(2,2)

The product of the digits written on cell X and cell Y is 1 for all of those pairs, so the answer is 5.

Sample Input 2

4
1111
11#1
1#11
1111

Sample Output 2

47

Sample Input 3

10
76##63##3#
8445669721
75#9542133
3#285##445
749632##89
2458##9515
5952578#77
1#3#44196#
4355#99#1#
#298#63587

Sample Output 3

36065

Sample Input 4

10
4177143673
7#########
5#1716155#
6#4#####5#
2#3#597#6#
6#9#8#3#5#
5#2#899#9#
1#6#####6#
6#5359657#
5#########

Sample Output 4

6525

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <string.h>
#include <inttypes.h>
#include <limits.h>
using namespace std;

using ll=int64_t;
//#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#ifdef MAROON_LOCAL
#define cerr (cerr<<"-- line "<<__LINE__<<" -- ")
#else
class CerrDummy{}cerrDummy;
template<class T>
CerrDummy& operator<<(CerrDummy&cd,const T&){
	return cd;
}
using charTDummy=char;
using traitsDummy=char_traits<charTDummy>;
CerrDummy& operator<<(CerrDummy&cd,basic_ostream<charTDummy,traitsDummy>&(basic_ostream<charTDummy,traitsDummy>&)){
	return cd;
}
#define cerr cerrDummy
#endif
#define REACH cerr<<"reached"<<endl
#define DMP(x) cerr<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))
#define ONE(x) memset(x,-1,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"{";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

ll read(){
	ll i;
	scanf("%" SCNd64,&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(ll x,int suc=1){
	printf("%" PRId64,x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341000];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341000];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(b<a)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

#define CAPITAL
void Yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<endl;
	#else
	cout<<"Yes"<<endl;
	#endif
	if(ex)exit(0);
}
void No(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<endl;
	#else
	cout<<"No"<<endl;
	#endif
	if(ex)exit(0);
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

const int Nmax=510;
int bot[Nmax][Nmax],l[Nmax],r[Nmax],imos[Nmax];

signed main(){
	int n=read();
	vector<string> g(n);
	REP(i,n)
		g[i]=readString();
	
	for(int i=n-1;i>=0;i--)for(int j=n-1;j>=0;j--){
		if(g[i][j]!='#'){
			bot[i][j]=i;
			if(i+1<n&&g[i+1][j]!='#')
				chmax(bot[i][j],bot[i+1][j]);
			if(j+1<n&&g[i][j+1]!='#')
				chmax(bot[i][j],bot[i][j+1]);
		}else
			bot[i][j]=-1;
	}
	
	ll ans=0;
	REP(top,n){
		FOR(i,top,n){
			REP(j,n)
				imos[j]=0;
			REP(j,n)
				if(g[i][j]!='#'){
					if(i==top){
						r[j]=j;
						l[j]=j;
					}
					if(j>0){
						chmax(r[j],r[j-1]);
						chmin(l[j],l[j-1]);
					}
					if(l[j]<=r[j]){
						int z=g[i][j]-'0';
						imos[l[j]]+=z;
						imos[r[j]+1]-=z;
					}
				}else{
					r[j]=-inf;
					l[j]=inf;
				}
			REP(j,n){
				if(bot[top][j]>=i)
					ans+=imos[j]*(g[top][j]-'0');
				imos[j+1]+=imos[j];
			}
		}
	}
	
	REP(i,n)REP(j,n)if(g[i][j]!='#')
		ans-=(g[i][j]-'0')*(g[i][j]-'0');
	
	print(ans);
}