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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n;
	string a[505];
	cin>>n;
	for(int i=0; i<n; i++){
		cin>>a[i];
	}
	int c1[505][505], c2[505][505], d[505][505];
	int s[505][505];
	ll ans=0;
	for(int i=0; i<n; i++){
		bool b[505][505]={};
		for(int j=0; j<n; j++) if(a[i][j]=='#') b[i][j]=1;
		for(int k=i+1; k<n; k++){
			for(int j=0; j<n; j++){
				if(a[k][j]=='#'){
					b[k][j]=1; continue;
				}
				if(b[k-1][j] && (j==0 || b[k][j-1])) b[k][j]=1;
			}
		}
		for(int k=i; k<n; k++){
			s[k][0]=0;
			for(int j=0; j<n; j++){
				s[k][j+1]=s[k][j];
				if(!b[k][j]) s[k][j+1]+=(a[k][j]-'0');
			}
		}
		for(int k=i; k<n; k++){
			for(int j=0; j<n; j++){
				if(b[k][j]){
					if(j>0) c1[k][j]=c1[k][j-1];
					else c1[k][j]=-1;
				}else c1[k][j]=j;
			}
			for(int j=n-1; j>=0; j--){
				if(b[k][j]){
					if(j<n-1) c2[k][j]=c2[k][j+1];
					else c2[k][j]=n;
				}else c2[k][j]=j;
				if(!b[k][j]){
					if(j==n-1 || b[k][j+1]) d[k][j]=j;
					else d[k][j]=d[k][j+1];
				}
			}
		}
		for(int j=0; j<n; j++){
			if(a[i][j]=='#') continue;
			int l=j, r=d[i][j];
			ll sum=s[i][r+1]-s[i][l+1];
			for(int k=i+1; k<n; k++){
				if(b[k][l]){
					l=c2[k][l];
				}
				if(b[k][r]){
					r=c1[k][r];
				}else{
					r=d[k][r];
				}
				if(l>r) break;
				sum+=s[k][r+1]-s[k][l];
			}
			ans+=sum*(a[i][j]-'0');
		}
	}
	cout<<ans<<endl;
	return 0;
}