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 cs const
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
cs int N = 505;
int n; char S[N][N];
int sm[N][N], dn[N][N]; 
int mn[2][N][N], mx[2][N][N];
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%s",S[i]+1);
 	int now=0, nxt=1; ll Ans=0;
 	for(int i=n,x; i; i--){
		for(int j=n; j; j--)
	 	if(S[i][j]!='#'){ 
	 		x=S[i][j]-'0';
	 		sm[i][j]=sm[i+1][j]+sm[i][j+1]+x;
	 		dn[i][j]=max(max(dn[i+1][j],dn[i][j+1]),i);
	 		mn[nxt][j][i]=j;
	 		for(int k=i+1; k<=dn[i+1][j]; k++) mn[nxt][j][k]=mn[now][j][k];
	 		for(int k=max(dn[i+1][j],i)+1; k<=dn[i][j+1]; k++) mn[nxt][j][k]=mn[nxt][j+1][k];
	 		mx[nxt][j][i]=j;
	 		for(int k=i; k<=dn[i][j+1]; k++) mx[nxt][j][k]=mx[nxt][j+1][k];
	 		for(int k=max(dn[i][j+1],i)+1; k<=dn[i+1][j]; k++) mx[nxt][j][k]=mx[now][j][k];
	 		int u=i+1, lm=min(dn[i][j+1],dn[i+1][j]);
	 		while(u<=lm){
	 			if(mn[nxt][j+1][u]<=mx[now][j][u]){
	 				sm[i][j]-=sm[u][mn[nxt][j+1][u]];
	 				u=dn[u][mn[nxt][j+1][u]];
	 			} ++u;
	 		} Ans+=(ll)x*(sm[i][j]-x);
		} swap(now,nxt);
	} cout<<Ans; return 0;
}