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 unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#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 CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=1505;
int n,cnt[N];
int a[N][N],S[N][N];
int L[N][N],R[N][N];
char s[N];
ll ans;
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]-'0');
		For(j,1,n) S[i][j]=S[i][j-1]+a[i][j];
	}
	For(i,1,n) Rep(j,n+1,1)
		if (!a[i][j]) L[i][j]=n+1,R[i][j]=0;
		else{
			L[i][j]=j;
			R[i][j]=max(R[i][j+1],j);
			ans+=a[i][j]*(S[i][R[i][j]]-S[i][j]); 
		}
	Rep(x,n,1) For(y,x+1,n){
		Rep(j,n,1)
			if (a[x][j]){
				L[y][j]=min(L[y][j],L[y][j+1]);
				R[y][j]=max(R[y][j],R[y][j+1]);
			}
			else{
				L[y][j]=n+1;
				R[y][j]=0;
			}
		For(j,1,n) cnt[j]=0;
		For(j,1,n)
			if (L[y][j]<=R[y][j]){
				ans+=a[x][j]*(S[y][R[y][j]]-S[y][L[y][j]-1]);
				++cnt[L[y][j]];
				--cnt[R[y][j]+1];
			}
		For(j,1,n){
			cnt[j]+=cnt[j-1];
			if (!cnt[j])
				a[y][j]=0;
		}
		For(j,1,n)
			S[y][j]=S[y][j-1]+a[y][j];
	}
	printf("%lld\n",ans);
}