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<cstdio>
#include<algorithm>
using namespace std;
char p[510][510];
int Up[510][510], Down[510][510];
int B[510][510], E[510][510], BB[510], EE[510], S[510];
int T[510][510], v[510];
int n, w[510][510], L[510], R[510];
long long res;
int main() {
	int i, j, k;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%s", p[i] + 1);
		for (j = 1; j <= n; j++) {
			if (p[i][j] >= '0' && p[i][j] <= '9')w[i][j] = p[i][j] - '0';
			else w[i][j] = -1;
		}
	}
	for (i = 0; i <= n + 1; i++)for (j = 0; j <= n + 1; j++)if (i == 0 || i > n || j == 0 || j > n)w[i][j] = -1;
	for (i = n; i >= 1; i--) {
		for (j = n; j >= 1; j--) {
			if (w[i][j] == -1)continue;
			Down[i][j] = i;
			Down[i][j] = max(Down[i][j], Down[i][j + 1]);
			Down[i][j] = max(Down[i][j], Down[i + 1][j]);
		}
	}
	for (i = 1; i <= n; i++) {
		for (j = 1; j <= n; j++) {
			if (w[i][j] == -1) {
				Up[i][j] = n + 1;
				continue;
			}
			Up[i][j] = i;
			if (w[i][j - 1] != -1)Up[i][j] = min(Up[i][j], Up[i][j - 1]);
			if (w[i - 1][j] != -1)Up[i][j] = min(Up[i][j], Up[i - 1][j]);
		}
	}
	for (i = 1; i <= n; i++) {
		int pv = 0, s = 0;
		for (j = 1; j <= n; j++) {
			if (w[i][j] == -1) {
				pv = j;
				B[i][j] = 1e9, E[i][j] = -1e9;
				s = 0;
				continue;
			}
			T[i][j] = pv + 1;
			res += s*w[i][j];
			s += w[i][j];
			int b = T[i][j], e = j;
			if (i != 1) {
				B[i][j] = 1e9, E[i][j] = -1e9;
				for (k = b; k <= e; k++) {
					if (w[i - 1][k] != -1) {
						B[i][j] = min(B[i][j], T[i - 1][k]);
						E[i][j] = max(E[i][j], k);
					}
				}
			}
		}
		for (j = 1; j <= n; j++) BB[j] = B[i][j], EE[j] = E[i][j];
		for (j = i - 1; j >= 1; j--) {
			for (k = 1; k <= n; k++) {
				S[k] = S[k - 1];
				if (Down[j][k] >= i) {
					S[k] += w[j][k];
				}
				if (j > 1) {
					if (Up[j][k] <= j-1 && Down[j][k] >= i)v[k] = 1;
					else v[k] = 0;
				}
			}
			for (k = 1; k <= n; k++) {
				if (Up[i][k] <= j) {
					res += (S[EE[k]] - S[BB[k] - 1])*w[i][k];
				}
			}
			if (j == 1)break;
			for (k = 1; k <= n; k++) {
				L[k] = L[k - 1];
				if (v[k])L[k] = k;
			}
			R[n + 1] = n + 1;
			for (k = n; k >= 1; k--) {
				R[k] = R[k + 1];
				if (v[k])R[k] = k;
			}
			for (k = 1; k <= n; k++) {
				int b = BB[k], e = EE[k];
				if (Up[i][k] >= j) {
					BB[k] = 1e8, EE[k] = -1e8;
					continue;
				}
				BB[k] = B[j][R[b]], EE[k] = E[j][L[e]];
			}
		}
	}
	printf("%lld\n", res);
}