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:
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const int MAXN = 2000;

int n;
string stmp[MAXN];
int a[MAXN][MAXN];
ll go[MAXN][MAXN];
int gd[MAXN][MAXN];
int gr[MAXN][MAXN];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n;
	for (int i = 0; i < n; ++i) {
		cin >> stmp[i];
		for (int j = 0; j < n; ++j) {
			if (stmp[i][j] == '#')
				a[i][j] = 0;
			else
				a[i][j] = stmp[i][j] - '0';
		}
	}
	ll ans = 0;
	for (int i = n - 1; i >= 0; --i)
		for (int j = n - 1; j >= 0; --j) {
			if (a[i][j] == 0)
				continue;
			if (i + 1 < n && j + 1 < n && a[i + 1][j] != 0 && a[i][j + 1] != 0) {
				go[i][j] += go[i + 1][j];
				go[i][j] += go[i][j + 1];
				//TODO: optimize
				for (int k = i + 1; k < n && gd[j + 1][k] != -1; ++k) {
					if (gr[j][gd[j + 1][k]] == k)
						go[i][j] -= go[k][gd[j + 1][k]];
				}
				gd[j][i] = j;
				for (int x = j + 1; x < n && gr[j + 1][x] != -1; ++x)
					gr[j][x] = gr[j + 1][x];
				int st = i + 1;
				while (st < n && gd[j][st] != -1)
					++st;
				while (st < n && gd[j + 1][st] != -1)
					gd[j][st] = gd[j + 1][st], ++st;
				gr[j][j] = i;
			}
			else if (i + 1 < n && a[i + 1][j] != 0) {
				go[i][j] += go[i + 1][j];
				gd[j][i] = j;
				gr[j][j] = i;
			}
			else if (j + 1 < n && a[i][j + 1] != 0) {
				go[i][j] += go[i][j + 1];
				memcpy(gr[j] + j + 1, gr[j + 1] + j + 1, sizeof(gr[0][0]) * (n - j - 1));
				memcpy(gd[j] + i + 1, gd[j + 1] + i + 1, sizeof(gr[0][0]) * (n - i - 1));
				gd[j][i] = j;
				gr[j][j] = i;
			}
			else {
				memset(gr[j] + j + 1, -1, sizeof(gr[0][0]) * (n - j - 1));
				memset(gd[j] + i + 1, -1, sizeof(gr[0][0]) * (n - i - 1));
				gd[j][i] = j;
				gr[j][j] = i;
			}
			ans += go[i][j] * a[i][j];
			go[i][j] += a[i][j];
		}
	cout << ans << "\n";
	return 0;
}