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>

using namespace std;

int main() {
#ifdef wxh010910
  freopen("input.txt", "r", stdin);
#endif
  int n;
  cin >> n;
  vector<string> board(n);
  for (int i = 0; i < n; ++i) {
    cin >> board[i];
  }
  vector<vector<int>> sum(n, vector<int> (n + 1));
  
  auto recalc_sum = [&](int x) {
    for (int i = 0; i < n; ++i) {
      sum[x][i + 1] = sum[x][i] + (board[x][i] == '#' ? 0 : board[x][i] - '0');
    }
  };

  auto calc_sum = [&](int x, int l, int r) {
    return sum[x][r + 1] - sum[x][l];
  };

  for (int i = 0; i < n; ++i) {
    recalc_sum(i);
  }

  vector<vector<int>> l(n, vector<int> (n, n)), r(n, vector<int> (n, -1));
  long long answer = 0;
  for (int x = n - 1; ~x; --x) {
    vector<vector<int>> tl(n, vector<int> (n, n)), tr(n, vector<int> (n, -1));
    for (int z = x; z < n; ++z) {
      if (z == x) {
        for (int y = n - 1; ~y; --y) {
          if (board[x][y] != '#') {
            tl[z][y] = y;
            tr[z][y] = max(y, y + 1 < n ? tr[z][y + 1] : -1);
            answer += (board[x][y] - '0') * calc_sum(z, tl[z][y] + 1, tr[z][y]);
          }
        }
      } else {
        for (int y = n - 1; ~y; --y) {
          if (board[x][y] != '#') {
            tl[z][y] = min(l[z][y], y + 1 < n ? tl[z][y + 1] : n);
            tr[z][y] = max(r[z][y], y + 1 < n ? tr[z][y + 1] : -1);
          }
        }
        vector<int> visit(n + 1);
        for (int y = 0; y < n; ++y) {
          if (tl[z][y] <= tr[z][y]) {
            ++visit[tl[z][y]];
            --visit[tr[z][y] + 1];
          }
        }
        for (int y = 0; y < n; ++y) {
          if (y) {
            visit[y] += visit[y - 1];
          }
          if (!visit[y]) {
            if (board[z][y] != '#') {
              board[z][y] = '#';
              recalc_sum(z);
            }
          }
        }
        for (int y = 0; y < n; ++y) {
          if (tl[z][y] <= tr[z][y]) {
            answer += (board[x][y] - '0') * calc_sum(z, tl[z][y], tr[z][y]);
          }
        }
      }
    }
    l = tl;
    r = tr;
  }

  cout << answer << endl;

  return 0;
}