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>

#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;

const int MX = 500, WORDS = (MX * MX + 63) >> 6;

pair<int, int> bit(int x, int y) {
	int f = max(MX - 1 - x, MX - 1 - y);
	int z = f * f;
	if (MX - 1 - x == f) z += MX - 1 - y;
	else z += f + 1 + MX - 1 - x;
	return {z >> 6, z & 63};
}

char s[MX][MX + 1];
uint64_t reach[MX + 1][WORDS];
uint64_t mask[WORDS][4];

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 0; i < n; i++) ignore = scanf(" %s", s[i]);
	//for (int i = 0; i < n; i++) fill(s[i], s[i] + n, '7');
	
	for (int x = 0; x < n; x++)
		for (int y = 0; y < n; y++) {
			if (s[x][y] == '#') continue;
			int a = s[x][y] - '0', p, q;
			tie(p, q) = bit(x, y);
			for (int k = 0; k < 4; k++) {
				mask[p][k] |= ((a >> k) & 1ULL) << q;
			}
		}
	
	long long ans = 0;
	for (int d = 2 * (n - 1); d >= 0; d--) {
		for (int x = 0, y = d; x <= d; x++, y--) {
			if (y >= n || x >= n || s[x][y] == '#') {
				if (x < n) memset(reach[x], 0, sizeof reach[x]);
				continue;
			}
			
			int sum0 = 0, sum1 = 0, sum2 = 0, sum3 = 0;
			uint64_t* r1 = reach[x];
			uint64_t* r2 = reach[x + 1];
			for (int i = 0; i < WORDS; i++) {
				r1[i] |= r2[i];
				if (r1[i] > 0) {
					sum0 += __builtin_popcountll(r1[i] & mask[i][0]);
					sum1 += __builtin_popcountll(r1[i] & mask[i][1]);
					sum2 += __builtin_popcountll(r1[i] & mask[i][2]);
					sum3 += __builtin_popcountll(r1[i] & mask[i][3]);
				}
			}
			
			ans += (sum0 + sum1 * 2 + sum2 * 4 + sum3 * 8) * (s[x][y] - '0');
			
			int p, q;
			tie(p, q) = bit(x, y);
			reach[x][p] |= (1ULL << q);
		}
	}
	
	printf("%lld\n", ans);
	
	return 0;
}