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>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
using namespace std;
int n,i,j,k,f[1505][1505],l[1505][1505],r[1505][1505];
long long ans;
char a[1505][1505];
int main(){
	cin>>n;
	rep(i,n){
		rep(j,n){
			cin>>a[i][j];
		}
	}
	for(i=0;i<=n+1;i++){
		a[0][i]=a[i][0]=a[n+1][i]=a[i][n+1]='#';
	}
	for(i=n;i;i--){
		for(j=n;j;j--){
			if(a[i][j]!='#'){
				if(a[i+1][j]=='#'&&a[i][j+1]=='#'){
					l[i][j]=r[i][j]=j;
					for(k=i+1;k<=n;k++){
						l[k][j]=n+1;
						r[k][j]=0;
					}
				}
				if(a[i+1][j]!='#'&&a[i][j+1]=='#'){
					f[i][j]=f[i+1][j];
					l[i][j]=r[i][j]=j;
				}
				if(a[i+1][j]=='#'&&a[i][j+1]!='#'){
					f[i][j]=f[i][j+1];
					l[i][j]=j;
					r[i][j]=r[i][j+1];
					for(k=i+1;k<=n;k++){
						l[k][j]=l[k][j+1];
						r[k][j]=r[k][j+1];
					}
				}
				if(a[i+1][j]!='#'&&a[i][j+1]!='#'){
					l[i][j]=j;r[i][j]=r[i][j+1];
					f[i][j]=f[i+1][j]+f[i][j+1];
					int lst=0;
					for(k=i+1;k<=n;k++){
						if(l[k][j+1]<=r[k][j]&&l[k][j+1]>lst){
							f[i][j]-=f[k][l[k][j+1]];
						}
						lst=max(lst,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]);
					}
				}
				ans+=f[i][j]*(a[i][j]&15);
				f[i][j]+=(a[i][j]&15);
			}
			else{
				l[i][j]=n+1;
				r[i][j]=0;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}