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:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)x.size())
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x){
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
const int N = 1505, inf = 0x3f3f3f3f;
int n;short l[N][N],r[N][N];
int dfn[N][N],pos[N][N];char s[N][N];
ll f[N][N],res;
#define umax(a,b) (a=a>b?a:b)
#define umin(a,b) (a=a<b?a:b)
int main() {
	memset(s,'#',sizeof(s));read(n);rep(i,1,n)scanf("%s",s[i]+1);
	memset(l,inf,sizeof(l));
	per(i,n,1){
		rep(j,1,n){
			if(s[i][j]!='#')l[i][j]=r[i][j]=j;else l[i][j]=inf,r[i][j]=0;
			l[i+1][j]=inf,r[i+1][j]=0;
		}
		per(j,i,1){
		short *R=r[j+1],*L=l[j+1];
		if(j<i)memcpy(r[j],r[j+1],sizeof(r[j+1])),memcpy(l[j],l[j+1],sizeof(l[j+1]));
		per(k,n,1)if(s[j][k]=='#')l[j][k]=inf,r[j][k]=0;
		else{
		//	if(j<i){r[j][k]=max(r[j][k+1],R[k]);l[j][k]=min(l[j][k+1],L[k]);}
		//	else
			{umax(r[j][k],r[j][k+1]);umin(l[j][k],l[j][k+1]);}
			if(dfn[j][k]==i+1&&R[k]<pos[j][k])f[j][k]-=f[i+1][pos[j][k]];
			if(R[k]>=l[j][k+1])dfn[j][k]=i,pos[j][k]=l[j][k+1];
		}
		}
		per(j,n,1)if(s[i][j]!='#'){
		//	printf("%d %d:%lld %lld\n",i,j,f[i][j],f[i+1][j]+f[i][j+1]+f[i][j]);
			f[i][j]+=f[i+1][j]+f[i][j+1];ll c=s[i][j]-'0';
			res+=c*f[i][j];f[i][j]+=c;
		}
	}
	cout<<res;return 0;
}