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:
//Author:xht37
#include <bits/stdc++.h>

#define ui unsigned int
#define ll long long
#define ul unsigned ll
#define ld long double

#define pi pair <int, int>
#define fi first
#define se second
#define mp make_pair

#define ls (p << 1)
#define rs (ls | 1)
#define md ((t[p].l + t[p].w) >> 1)

#define vi vector <int>
#define pb push_back
#define pq priority_queue

#define dbg(x) cerr << #x" = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)

#define fl(x) freopen(x".in", "w", stdin), freopen(x".out", "w", stdout)

using namespace std;

namespace io {
	const int SI = 1 << 21 | 1;
	char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
	int f, t;
	#define gc() (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin), IS == IT ? EOF : *IS++) : *IS++)
	inline void flush() {
		fwrite(OB, 1, OS - OB, stdout), OS = OB;
	}
	inline void pc(char x) {
		*OS++ = x;
		if (OS == OT) flush();
	}

	template <class I>
	inline void rd(I &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15), c = gc());
		x *= f;
	}
	template <class I>
	inline void rd(I &x, I &y) {
		rd(x), rd(y);
	}
	template <class I>
	inline void rd(I &x, I &y, I &z) {
		rd(x), rd(y), rd(z);
	}
	template <class I>
	inline void rda(I *a, int n) {
		for (int i = 1; i <= n; i++) rd(a[i]);
	}
	inline void rdc(char &c) {
		for (c = gc(); c < 33 || c > 126; c = gc());
	}
	inline void rds(char *a, int &n) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (n = 0; c >= 33 && c <= 126; a[++n] = c, c = gc());
		a[n+1] = '\0';
	}
	inline void rds(string &a) {
		for (c = gc(); c < 33 || c > 126; c = gc());
		for (a.clear(); c >= 33 && c <= 126; a.pb(c), c = gc());
	}

	template <class I>
	inline void print(I x, char k = '\n') {
		if (!x) pc('0');
		if (x < 0) pc('-'), x = -x;
		while (x) ch[++t] = x % 10 + '0', x /= 10;
		while (t) pc(ch[t--]);
		pc(k);
	}
	template <class I>
	inline void print(I x, I y) {
		print(x, ' '), print(y);
	}
	template <class I>
	inline void print(I x, I y, I z) {
		print(x, ' '), print(y, ' '), print(z);
	}
	template <class I>
	inline void printa(I *a, int n) {
		for (int i = 1; i <= n; i++) print(a[i], " \n"[i==n]);
	}
	inline void printc(char c) {
		pc(c);
	}
	inline void prints(char *a, int n) {
		for (int i = 1; i <= n; i++) pc(a[i]);
		pc('\n');
	}
	inline void prints(string a) {
		int n = a.length();
		while (t < n) pc(a[t++]);
		pc('\n'), t = 0;
	}
	struct Flush {
		~Flush() {
			flush();
		}
	} flusher;
}
using io::rd;
using io::rda;
using io::rdc;
using io::rds;
using io::print;
using io::printa;
using io::printc;
using io::prints;

const int N = 1.5e3 + 7;
int n, s[N][N], p[N][N], l[2][N][N], r[2][N][N];
char a[N][N];
ll ans;

int main() {
	rd(n);
	for (int i = 1; i <= n; i++) rds(a[i], n);
	for (int i = n, o = 0; i; i--, o ^= 1)
		for (int j = n; j > 0; j--)
			if (a[i][j] != '#') {
				a[i][j] -= '0';
				s[i][j] = s[i][j+1] + s[i+1][j] + a[i][j];
				p[i][j] = max(max(p[i][j+1], p[i+1][j]), i);
				l[o][j][i] = j;
				for (int k = i + 1; k <= p[i+1][j]; k++)
					l[o][j][k] = l[o^1][j][k];
				for (int k = max(p[i+1][j], i) + 1; k <= p[i][j+1]; k++)
					l[o][j][k] = l[o][j+1][k];
				r[o][j][i] = j;
				for (int k = i; k <= p[i][j+1]; k++)
					r[o][j][k] = r[o][j+1][k];
				for (int k = max(p[i][j+1], i) + 1; k <= p[i+1][j]; k++)
					r[o][j][k] = r[o^1][j][k];
				for (int k = i + 1; k <= min(p[i][j+1], p[i+1][j]); k++)
					if (r[o^1][j][k] >= l[o][j+1][k])
						s[i][j] -= s[k][l[o][j+1][k]],
						k = p[k][l[o][j+1][k]];
				ans += 1ll * a[i][j] * (s[i][j] - a[i][j]);
			}
	print(ans);
	return 0;
}