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>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	typedef long long ll;
	const int N=509;
	ll ans;int n,maxn[2][N][N],minn[2][N][N],sum[N][N],pos[N][N];char s[N][N];
	int main(){
		n=read<int>();
		for(int i=1;i<=n;i++) scanf("%s",s[i]+1);

		int cur=0;
		for(int i=n;i;i--){
			int pre=cur;cur^=1;
			for(int j=n;j;j--) if(s[i][j]!='#'){
				sum[i][j]=sum[i][j+1]+sum[i+1][j]+s[i][j]-'0';
				pos[i][j]=max(max(pos[i][j+1],pos[i+1][j]),i);

				minn[cur][j][i]=j;
				for(int k=i+1;k<=pos[i+1][j];k++) minn[cur][j][k]=minn[pre][j][k];
				for(int k=max(pos[i+1][j],i)+1;k<=pos[i][j+1];k++) minn[cur][j][k]=minn[cur][j+1][k];
				
				maxn[cur][j][i]=j;
				for(int k=i;k<=pos[i][j+1];k++) maxn[cur][j][k]=maxn[cur][j+1][k];
				for(int k=max(pos[i][j+1],i)+1;k<=pos[i+1][j];k++) maxn[cur][j][k]=maxn[pre][j][k];

				int d=i+1,r=min(pos[i+1][j],pos[i][j+1]);
				while(d<=r){
					if(maxn[pre][j][d]>=minn[cur][j+1][d]){
						sum[i][j]-=sum[d][minn[cur][j+1][d]];
						d=pos[d][minn[cur][j+1][d]];
					}
					d++;
				}
				ans+=1LL*(s[i][j]-'0')*(sum[i][j]-s[i][j]+'0');
			}
		}
		printf("%lld\n",ans);
		return 0;
	}
}
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}