Problem: 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.

Examples

Input

2
11
11


Output

5


Input

4
1111
11#1
1#11
1111


Output

47


Input

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


Output

36065


Input

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#########


Output

6525

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = 1503;
int n;
char s[N][N];
int dp[N][N];
int leftmost[2][N][N], upmost[2][N][N];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d", &n);
	for (int i = 0; i < n; i++)
		scanf("%s", s[i]);
	for (int i = 0; i <= n; i++)
		s[n][i] = s[i][n] = '#';
	ll ans = 0;
	for (int x = n - 1; x >= 0; x--) {
		for (int y = n - 1; y >= 0; y--) {
			for (int k = 0; k < n; k++)
				leftmost[0][y][k] = upmost[0][y][k] = N;
			if (s[x][y] == '#') continue;
			if (s[x + 1][y] != '#')
				for (int k = 0; k < n; k++) {
					leftmost[0][y][k] = min(leftmost[0][y][k], leftmost[1][y][k]);
					upmost[0][y][k] = min(upmost[0][y][k], upmost[1][y][k]);
				}
			if (s[x][y + 1] != '#')
				for (int k = 0; k < n; k++) {
					leftmost[0][y][k] = min(leftmost[0][y][k], leftmost[0][y + 1][k]);
					upmost[0][y][k] = min(upmost[0][y][k], upmost[0][y + 1][k]);
				}
			leftmost[0][y][x] = y;
			upmost[0][y][y] = x;
			
			//eprintf("cell = %d %d\n", x, y);
			/*
			for (int k = 0; k < n; k++)
				eprintf("%d ", leftmost[0][y][k]);
			eprintf("\n");
			for (int k = 0; k < n; k++)
				eprintf("%d ", upmost[0][y][k]);
			eprintf("\n");
			*/
			dp[x][y] = dp[x + 1][y] + dp[x][y + 1];
			if (x != n - 1 && y != n - 1) {
				for (int xx = x + 1; xx < n; xx++) {
					int yy = leftmost[0][y + 1][xx];
					//eprintf("!!! %d %d\n", xx, yy);
					if (upmost[1][y][yy] == xx)
						dp[x][y] -= dp[xx][yy];
				}
			}
			int c = (int)(s[x][y] - '0');
			ans += c * dp[x][y];
			dp[x][y] += c;
		}
		for (int y = 0; y < n; y++)
			for (int k = 0; k < n; k++) {
				leftmost[1][y][k] = leftmost[0][y][k];
				upmost[1][y][k] = upmost[0][y][k];
			}
	}
	/*
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++)
			printf("%d ", dp[i][j]);
		printf("\n");
	}
	*/
	printf("%lld\n", ans);

	return 0;
}