Problem:
Score : 1500 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 1500

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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
public class Main {

    int[][] cells;
    int ans;
    boolean[][] already;
    
    public static void main(String[] args) throws IOException {
	Main m = new Main();
	m.solve();
    }

    private void solve() throws IOException {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	int n = Integer.parseInt(br.readLine());
	cells = new int[n+1][n+1];
	for (int i=0;i<n;i++) {
	    cells[i][n] = cells[n][i] = -1;
	    String[] a = br.readLine().split("");
	    for (int j=0;j<n;j++) {
		if (a[j].equals("#")) {
		    cells[i][j] = -1;
		    continue;
		}
		cells[i][j] = Integer.parseInt(a[j]);
	    }
	}
	br.close();
	ans = 0;
	for (int i=0;i<n;i++) {
	    for (int j=0;j<n;j++) {
		if (cells[i][j] < 0) continue;
		//System.out.println("from "+i+" "+j);
		already = new boolean[n+1][n+1];
		search(i,j,cells[i][j]);
	    }
	}
	System.out.println(ans);
    }

    private void search(int i,int j,int c) {
	if (cells[i+1][j] > 0 && !already[i+1][j]) {
	    //System.out.println("at "+(i+1)+" "+j);
	    already[i+1][j] = true;
	    ans += c*cells[i+1][j];
	    search(i+1,j,c);
	}
	if (cells[i][j+1] > 0) {
	    //System.out.println("at "+i+" "+(j+1));
	    already[i][j+1] = true;
	    ans += c*cells[i][j+1];
	    search(i,j+1,c);
	}
    }
}