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<algorithm>
#include<cstring>
#include<cmath>

using namespace std;
typedef long long ll;
int n;
int a[1505][1505];
char ss[1505];
int L[1505][1505],R[1505][1505];
ll f[1505][1505];
ll ans;
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%s",&ss[1]);
		for(int j=1;j<=n;j++) {
			a[i][j] = (ss[j]=='#'?0:ss[j]-'0');
		}
	}
	for(int i=n;i>=1;i--) {
		for(int j=n;j>=1;j--) {
			if(a[i][j]) {
				bool mk1 = a[i+1][j] , mk2 = a[i][j+1];
				if( (!mk1) && (!mk2) ) {
					L[i][j] = R[i][j] = j;
					f[i][j] = 0;
					for(int k=i+1;k<=n;k++) L[k][j] = n+1,R[k][j] = 0;
				} else if(!mk1) {
					f[i][j] = f[i][j+1];
					L[i][j] = j; R[i][j] = R[i][j+1];
					for(int k=i+1;k<=n;k++) L[k][j] = L[k][j+1] , R[k][j] = R[k][j+1];
				} else if(!mk2) {
					f[i][j] = f[i+1][j];
					L[i][j] = R[i][j] = j;
				} else {
					int msr = 0;
					int yoo = f[i][j+1] + f[i+1][j];
					L[i][j] = j; R[i][j] = R[i][j+1];
					for(int k=i+1;k<=n;k++) {
						if( (L[k][j+1]>msr) && (L[k][j+1]<=R[k][j])  ) yoo -= f[k][L[k][j+1]];
						msr = max(msr,R[k][j]);
						L[k][j] = min(L[k][j],L[k][j+1]);
						R[k][j] = max(R[k][j],R[k][j+1]);
					}
					f[i][j] = yoo;
				}
				ans += f[i][j] * a[i][j];
				f[i][j] += a[i][j];
			} else L[i][j] = n+1,R[i][j] = 0;
		}
	}
	printf("%lld",ans);
}