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 ll long long
#define ull unsigned ll
#define uint ungigned
#define db double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pli pair<ll,int>
#define vi vector<int>
#define vpi vector<pii >
#define IT iterator
 
#define PB push_back
#define MK make_pair
#define LB lower_bound
#define UB upper_bound
#define y1 wzpakking 
#define fi first
#define se second
#define BG begin
#define ED end
 
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define UPD(x,y) (((x)+=(y))>=mo?(x)-=mo:233)
#define CLR(a,v) memset(a,v,sizeof(a))
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define sqr(x) (1ll*x*x)
 
#define LS3 k*2,l,mid
#define RS3 k*2+1,mid+1,r
#define LS5 k*2,l,mid,x,y
#define RS5 k*2+1,mid+1,r,x,y
#define GET pushdown(k);int mid=(l+r)/2
#define INF (1<<29)
using namespace std;
const int N=505;
char s[N];
int n,cnt[N];
ll ans;
int a[N][N],sum[N][N];
int l[N][N],r[N][N];
int tl[N][N],tr[N][N];
void recalc(int i){
	For(j,1,n) sum[i][j]=sum[i][j-1]+a[i][j];
}
int S(int i,int l,int r){
	return sum[i][r]-(l?sum[i][l-1]:0);
}
int main(){
	scanf("%d",&n);
	For(i,1,n){
		scanf("%s",s+1);
		For(j,1,n) a[i][j]=(s[j]=='#'?0:s[j]-48);
		recalc(i);
	}
	For(i,1,n) For(j,1,n)
		l[i][j]=n+1,r[i][j]=0;
	Rep(x,n,1){
		For(i,1,n) For(j,1,n)
			tl[i][j]=n+1,tr[i][j]=0;
		For(y,x,n)
			if (y==x){
				Rep(j,n,1) if(a[y][j]){
					tl[y][j]=j;
					tr[y][j]=max(j,j==n?0:tr[y][j+1]);
					ans+=1ll*a[y][j]*S(y,tl[y][j]+1,tr[y][j]);
				}
			}
			else{
				Rep(j,n,1) if (a[x][j]){
					tl[y][j]=min(l[y][j],j==n?n+1:tl[y][j+1]);
					tr[y][j]=max(r[y][j],j==n?0:tr[y][j+1]);
				}
				For(j,1,n) cnt[j]=0;
				For(j,1,n) if (tl[y][j]<=tr[y][j]){
					++cnt[tl[y][j]];
					--cnt[tr[y][j]+1];
				}
				For(j,1,n){
					cnt[j]+=cnt[j-1];
					if (!cnt[j]&&a[y][j])
						a[y][j]=0,recalc(y);
				}
				For(j,1,n)
					if (tl[y][j]<=tr[y][j])
						ans+=1ll*a[x][j]*S(y,tl[y][j],tr[y][j]);
			}
		For(i,1,n) For(j,1,n)
			l[i][j]=tl[i][j],r[i][j]=tr[i][j];
	}
	printf("%lld",ans);
}