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<bits/stdc++.h>
using namespace std;
const int MAXN = 1505;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
char s[MAXN][MAXN]; int n, a[MAXN][MAXN];
int l[MAXN][MAXN], r[MAXN][MAXN], res[MAXN][MAXN];
int main() {
	read(n);
	for (int i = 1; i <= n; i++) {
		scanf("%s", s[i] + 1);
		for (int j = 1; j <= n; j++)
			if (s[i][j] == '#') a[i][j] = 0;
			else a[i][j] = s[i][j] - '0';
	}
	ll finalans = 0;
	for (int i = n; i >= 1; i--)
	for (int j = n; j >= 1; j--) {
		if (a[i][j] != 0) {
			if (a[i + 1][j] == 0 && a[i][j + 1] == 0) l[j][i] = r[j][i] = j;
			else if (a[i + 1][j] == 0) {
				for (int k = i; k <= n; k++) {
					l[j][k] = l[j + 1][k];
					r[j][k] = r[j + 1][k];
				} l[j][i] = j;
				res[i][j] = res[i][j + 1];
			} else if (a[i][j + 1] == 0) {
				l[j][i] = r[j][i] = j;
				res[i][j] = res[i + 1][j];
			} else {
				int Max = 0, ans = res[i + 1][j] + res[i][j + 1];
				l[j][i] = j, r[j][i] = r[j + 1][i];
				for (int k = i + 1; k <= n; k++) {
					if (r[j][k] >= l[j + 1][k] && l[j + 1][k] > Max) ans -= res[k][l[j + 1][k]];
					chkmax(Max, r[j][k]);
					chkmax(r[j][k], r[j + 1][k]);
					chkmin(l[j][k], l[j + 1][k]);
				}
				res[i][j] = ans;
			}
			finalans += a[i][j] * res[i][j];
			res[i][j] += a[i][j];
		} else {
			for (int k = i; k <= n; k++)
				l[j][k] = n + 1, r[j][k] = 0;
		}
	}
	cout << finalans << endl;
	return 0;
}