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;

typedef long long lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;
	
	char buf[bufl],*s=buf,*t=buf;
	
	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}
	
	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}

	inline int tc()
	{
		int c=fetch();
		while(c<=32 && c!=EOF)c=fetch();
		return c;
	}
}
using ae86::ty;
using ae86::tc;

const int _ = 1503 , inf = 0x3f3f3f3f;

int n,val[_][_],sv[_][_]={0},xl[_][_],xr[_][_];

int main()
{
	n=ty();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		{
			val[i][j]=tc();
			if(!isdigit(val[i][j]))val[i][j]=0;
			else val[i][j]-=48;
		}

	lint ans=0;
	for(int i=n;i>=1;i--)
		for(int j=n;j>=1;j--)
		{
			if(val[i][j])
			{
				if(val[i+1][j] && val[i][j+1])
				{
					int temp=sv[i+1][j]+sv[i][j+1],nowr=0;
					xl[j][i]=j,xr[j][i]=xr[j+1][i];
					for(int k=i+1;k<=n;k++)
					{
						if(xr[j][k]>=xl[j+1][k] && xl[j+1][k]>nowr)temp-=sv[k][xl[j+1][k]];
						nowr=max(nowr,xr[j][k]);
						xl[j][k]=min(xl[j][k],xl[j+1][k]),xr[j][k]=max(xr[j][k],xr[j+1][k]);
					}
					sv[i][j]=temp;	
				}
				else if(val[i+1][j])
				{
					xl[j][i]=xr[j][i]=j,sv[i][j]=sv[i+1][j];
				}
				else if(val[i][j+1])
				{
					for(int k=i;k<=n;k++)xl[j][k]=xl[j+1][k],xr[j][k]=xr[j+1][k];
					xl[j][i]=j,sv[i][j]=sv[i][j+1];
				}
				else xl[j][i]=xr[j][i]=j;
				ans+=1ll*sv[i][j]*val[i][j],sv[i][j]+=val[i][j];
			}
			else for(int k=i;k<=n;k++)xl[j][k]=inf,xr[j][k]=-inf;
		}

	printf("%lld\n",ans);

	return 0;
}