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;

template <typename T> void chmin(T&x,const T &y)
{
	if(x>y)x=y;
}
template <typename T> void chmax(T &x,const T &y)
{
	if(x<y)x=y;
}
typedef long long s64;
typedef unsigned long long u64;
typedef unsigned int u32;
typedef pair<int,int> pii;
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define per(i,r,l) for(int i=r;i>=l;--i)
#define rep0(i,l,r) for(int i=l;i<r;++i)
#define gc (c=getchar())
int read()
{
	char c;
	while(gc<'-');
	if(c=='-')
	{
		int x=gc-'0';
		while(gc>='0')x=x*10+c-'0';
		return -x;
	}
	int x=c-'0';
	while(gc>='0')x=x*10+c-'0';
	return x;
}
#undef gc

const int N=500+5;
char a[N][N];
int dp[N][N];
int cl[N][N],cr[N][N];

int main()
{
#ifdef kcz
	freopen("1.in","r",stdin);//freopen("1.out","w",stdout);
#endif
	int n;
	cin>>n;
	rep(i,1,n)scanf("%s",a[i]+1);
	s64 ans=0;
	per(j,n,1)
	if(isdigit(a[n][j]))
	{
		cl[j][n]=cr[j][n]=j;
		if(isdigit(a[n][j+1]))
		{
			cr[j][n]=cr[j+1][n];
			dp[n][j]=dp[n][j+1];
		}
		ans+=(a[n][j]-'0')*dp[n][j];
		dp[n][j]+=a[n][j]-'0';
	}
	per(i,n-1,1)
	per(j,n,1)
	if(isdigit(a[i][j]))
	{
		bool b1=isdigit(a[i][j+1]),b2=isdigit(a[i+1][j]);
		if(!b1&&!b2)
		{
			cl[j][i]=cr[j][i]=j;
			rep(k,i+1,n){cl[j][k]=n+1;cr[j][k]=0;}
		}else
		if(!b1&&b2)
		{
			dp[i][j]=dp[i+1][j];
			cl[j][i]=cr[j][i]=j;
		}else
		if(b1&&!b2)
		{
			dp[i][j]=dp[i][j+1];
			cl[j][i]=j;cr[j][i]=cr[j+1][i];
			rep(k,i+1,n){cl[j][k]=cl[j+1][k];cr[j][k]=cr[j+1][k];}
		}else
		{
			if(i==1&&j==2)
				int yyx=1;
			int ndp=dp[i+1][j]+dp[i][j+1];
			int last=0;
			cl[j][i]=j;cr[j][i]=cr[j+1][i];
			rep(k,i+1,n)
			{
				if(cr[j][k]>=cl[j+1][k]&&last<cl[j+1][k])
					ndp-=dp[k][cl[j+1][k]];
				last=cr[j][k];
				chmin(cl[j][k],cl[j+1][k]);
				chmax(cr[j][k],cr[j+1][k]);
			}
			dp[i][j]=ndp;
		}
		ans+=(a[i][j]-'0')*dp[i][j];
		dp[i][j]+=a[i][j]-'0';
	}
//	rep(i,1,n)
//rep(j,1,n)printf("%d%c",dp[i][j]," \n"[j==n]);
	cout<<ans;
}