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 <iostream>
#include <algorithm>
using namespace std;
#define ll long long
const int Q=1505;
char a[Q][Q];
int f[Q][Q],l[Q][Q],r[Q][Q];
int n;
int Ok(int x,int y)
{return (x<=n)&(y<=n)&(a[x][y]!='#');}
int main()
{
    ll als=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%s",a[i]+1);
    for(int i=n;i;--i)
        for(int j=n;j;--j){
            if(!Ok(i,j)){
                l[i][j]=n+1,r[i][j]=-1;
                continue;
            }
            int D=Ok(i+1,j),R=Ok(i,j+1);
            int val=a[i][j]-'0';
            if(!(D|R)){
                f[i][j]=val;
                l[i][j]=r[i][j]=j;
                for(int k=i+1;k<=n;k++)
                    l[k][j]=n+1,r[k][j]=0;
            }
            else if(D&R){
                f[i][j]=f[i+1][j]+f[i][j+1]+val;
                for(int k=i+1,mxr=0;k<=n;k++){
                    int &lv=l[k][j+1],&rv=r[k][j+1];
                    if(lv>mxr&&lv<=r[k][j])f[i][j]-=f[k][lv];
                    mxr=max(mxr,r[k][j]);
                    l[k][j]=min(l[k][j],lv);
                    r[k][j]=max(r[k][j],rv);
                }
                l[i][j]=j;
                r[i][j]=r[i][j+1];
            }
            else if(D){
                f[i][j]=f[i+1][j]+val;
                l[i][j]=r[i][j]=j;
            }
            else{
                f[i][j]=f[i][j+1]+val;
                
                for(int k=i;k<=n;k++)
                    l[k][j]=l[k][j+1],r[k][j]=r[k][j+1];l[i][j]=j;
            }
            als+=1LL*val*(f[i][j]-val);
        }
    printf("%lld\n",als);
    return 0;
}