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
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int sumv[505][505],rpos[505][505];
int minn[2][505][505],maxn[2][505][505];

char str[505][505];

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) scanf("%s",str[i]+1);
  int cur=0;
  ll ans=0;
  for(int i=n;i>0;i--) {
  	cur^=1;
  	for(int j=n;j>0;j--)
  	  if (str[i][j]!='#') {
  	  	  sumv[i][j]=sumv[i][j+1]+sumv[i+1][j]+str[i][j]-'0';
  	  	  rpos[i][j]=max(max(rpos[i][j+1],rpos[i+1][j]),i);
  	  	  minn[cur][j][i]=j;
  	  	  for(int k=i+1;k<=rpos[i+1][j];k++) minn[cur][j][k]=minn[cur^1][j][k];
  	  	  for(int k=max(rpos[i+1][j],i)+1;k<=rpos[i][j+1];k++) minn[cur][j][k]=minn[cur][j+1][k];
  	  	  maxn[cur][j][i]=j;
  	  	  for(int k=i;k<=rpos[i][j+1];k++) maxn[cur][j][k]=maxn[cur][j+1][k];
  	  	  for(int k=max(rpos[i][j+1],i)+1;k<=rpos[i+1][j];k++) maxn[cur][j][k]=maxn[cur^1][j][k];
  	  	  int d=i+1,r=min(rpos[i][j+1],rpos[i+1][j]);
  	  	  while (d<=r) {
  	  	  	    if (maxn[cur^1][j][d]>=minn[cur][j+1][d]) {
  	  	  	    	    sumv[i][j]-=sumv[d][minn[cur][j+1][d]];
  	  	  	    	    d=rpos[d][minn[cur][j+1][d]];
					  }
  	  	  	    d++;
			  }
		  ans+=(ll)(str[i][j]-'0')*(sumv[i][j]-(str[i][j]-'0'));
		}
  }
  printf("%lld\n",ans);
  return 0;
}