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: Time Limit Exceeded)

#include <iostream>
#include <vector>

bool CanReach(const std::vector<std::vector<char>>& board, int x1, int y1, int x2, int y2) {
	if (x1 > x2 || y1 > y2)return false;
	if (x1 == x2 && y1 == y2)return true;
	if (board[y1][x1] == '#')return false;
	if (!CanReach(board, x1 + 1, y1, x2, y2))return CanReach(board, x1, y1 + 1, x2, y2);
	return true;
}

int main() {

	int N;
	std::cin >> N;
	std::vector<std::vector<char>> board(N, std::vector<char>(N));
	for (auto&v : board) {
		getchar();
		for (auto&w : v)w = getchar();
	}
	int sum = 0;
	for (int x = 0; x < N; ++x) {
		for (int y = 0; y < N; ++y) {
			if (board[y][x] != '#') {
				for (int x2 = x, y2 = y + 1; x2 < N; ++x2, y2 = 0) {
					for (; y2 < N; ++y2) {
						if (board[y2][x2] != '#') {
							if (CanReach(board, x, y, x2, y2)) {
								sum += (board[y][x] - '0')*(board[y2][x2] - '0');
							}
						}
					}
				}
			}
		}
	}

	std::cout << sum << std::endl;

	return 0;
}