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: Wrong Answer)

#include <bits/stdc++.h>
#define for1(a,b,i) for(int i=a;i<=b;++i)
#define FOR2(a,b,i) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;

#define M 2005
int n;
char s[M];
int a[M][M],f[M][M],l[M][M],r[M][M];

int main () {
//	freopen("a.in","r",stdin);
	scanf("%d",&n);
	for1(1,n,i) {
		scanf("%s",s+1);
		for1(1,n,j) a[i][j]= s[j]=='#'?0:s[j]-'0';
	}
	
	ll ans=0;
	FOR2(n,1,i) FOR2(n,1,j) if(a[i][j]) {
		bool t[2]={a[i+1][j],a[i][j+1]};
		if(!(t[0]|t[1])) {
			l[i][j]=r[i][j]=j;
			for1(i+1,n,k) l[k][j]=n+1,r[k][j]=0;
		}
		else if(t[0]&&!t[1]) {
			l[i][j]=r[i][j]=j;
			f[i][j]=f[i+1][j];
		}
		else if(!t[0]&&t[1]) {
			f[i][j]=f[i][j+1];
			l[i][j]=j,r[i][j]=r[i][j+1];
			for1(i+1,n,k) l[k][j]=l[k][j+1],r[k][j]=r[k][j+1];
		}
		else {
			f[i][j]=f[i+1][j]+f[i][j+1];
			l[i][j]=j,r[i][j]=r[i][j+1];
			
			int las_r=0;		
			for1(i+1,n,k) {
				if(r[k][j]>=l[k][j+1]&&l[k][j+1]>las_r) {
					f[i][j]-=f[k][l[k][j+1]];
			//		cout<<i<<" "<<j<<" "<<k<<" "<<l[k][j+1]<<endl;
					las_r=r[k][j];
				}
				if(l[k][j]>r[k][j]) l[k][j]=l[k][j+1],r[k][j]=r[k][j+1];
			//	if(i==1&&j==1) 
			//cout<<l[k][j]<<" "<<r[k][j]<<" "<<l[k][j+1]<<" "<<r[k][j+1]<<endl;
			}
		}
		
//		cout<<i<<" "<<j<<" "<<f[i][j]<<endl;
		ans+=a[i][j]*f[i][j];
		f[i][j]+=a[i][j];
	}
	cout<<ans<<endl;
}