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 title "cell"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<ll,ll>
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
ll read(){
	ll g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const ll N=1505;
char s[N][N];
ll n,su[N][N],minn[2][N][N],maxn[2][N][N],mxto[N][N],ans;
signed main(){
	n=read();
	for(ll i=1;i<=n;i++)scanf("%s",s[i]+1);
	for(ll i=n,now=1,pre=0;i>=1;i--,swap(now,pre)){
		for(ll j=n;j>=1;j--)if(s[i][j]!='#'){
			su[i][j]=su[i][j+1]+su[i+1][j]+s[i][j]-'0';
			mxto[i][j]=max(max(mxto[i][j+1],mxto[i+1][j]),i);
			minn[now][j][i]=maxn[now][j][i]=j;
			for(ll k=i+1;k<=mxto[i+1][j];k++)minn[now][j][k]=minn[pre][j][k];
			for(ll k=max(mxto[i+1][j],i)+1;k<=mxto[i][j+1];k++)minn[now][j][k]=minn[now][j+1][k];
			for(ll k=i;k<=mxto[i][j+1];k++)maxn[now][j][k]=maxn[now][j+1][k];
			for(ll k=max(mxto[i][j+1],i)+1;k<=mxto[i+1][j];k++)maxn[now][j][k]=maxn[pre][j][k];
			ll pos=i+1,lim=min(mxto[i][j+1],mxto[i+1][j]);
			while(pos<=lim){
				if(maxn[pre][j][pos]>=minn[now][j+1][pos]){
					su[i][j]-=su[pos][minn[now][j+1][pos]];
					pos=mxto[pos][minn[now][j+1][pos]];
				}pos++;
			}ans+=(s[i][j]-'0')*(su[i][j]-(s[i][j]-'0'));
		}
	}return cout<<ans,signed();
}