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:  tourist
 *    created: 13.10.2018 16:23:29       
**/                        
#pragma GCC optimize("-Ofast")

#undef _GLIBCXX_DEBUG
 
#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 1505;
 
short L[2][N][N];
short R[2][N][N];
int a[N][N];
int row[N][N];
int up[N][N];
string board[N];
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> board(n);
  for (int i = 0; i < n; i++) {
    cin >> board[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      a[i][j] = (board[i][j] == '#' ? 0 : (int) (board[i][j] - '0'));
    }
  }
  for (int i = 0; i < n; i++) {
    row[i][0] = 0;
    for (int j = 0; j < n; j++) {
      row[i][j + 1] = row[i][j] + a[i][j];
    }
  }
  long long ans = 0;
  function<void(int,int)> put = [&](int i, int j) {
    if (0 <= i && i < n && 0 <= j && j < n && a[i][j] > 0) {
      if (i == 0 || a[i - 1][j] == 0) {
        if (j == 0 || a[i][j - 1] == 0) {
          for (int k = j; k < n; k++) {
            row[i][k + 1] -= a[i][j];
          }
          a[i][j] = 0;
          put(i + 1, j);
          put(i, j + 1);
        }
      }
    }
  };
  for (int i = n - 1; i >= 0; i--) {
    for (int j = n - 1; j >= 0; j--) {
      if (a[i][j] == 0) {
        continue;
      }
      int sum = 0;
      if (i < n - 1 && a[i + 1][j] != 0 && j < n - 1 && a[i][j + 1] != 0) {
        L[i & 1][j][i] = j;
        R[i & 1][j][i] = R[i & 1][j + 1][i];
        sum += row[i][R[i & 1][j][i] + 1] - row[i][L[i & 1][j][i]];
        up[i][j] = max(up[i + 1][j], up[i][j + 1]);
        int to = min(up[i + 1][j], up[i][j + 1]);
        memcpy(L[i & 1][j] + (i + 1), L[(i + 1) & 1][j] + (i + 1), sizeof(short) * (to - (i + 1) + 1));
        memcpy(R[i & 1][j] + (i + 1), R[i & 1][j + 1] + (i + 1), sizeof(short) * (to - (i + 1) + 1));
        for (int k = i + 1; k <= to; k++) {
          sum += row[k][R[i & 1][j][k] + 1] - row[k][L[i & 1][j][k]];
        }
        if (up[i + 1][j] < up[i][j + 1]) {
          memcpy(L[i & 1][j] + (to + 1), L[i & 1][j + 1] + (to + 1), sizeof(short) * (up[i][j] - (to + 1) + 1));
          memcpy(R[i & 1][j] + (to + 1), R[i & 1][j + 1] + (to + 1), sizeof(short) * (up[i][j] - (to + 1) + 1));
          for (int k = to + 1; k <= up[i][j]; k++) {
            sum += row[k][R[i & 1][j][k] + 1] - row[k][L[i & 1][j][k]];
          }
        }
        if (up[i + 1][j] > up[i][j + 1]) {
          memcpy(L[i & 1][j] + (to + 1), L[(i + 1) & 1][j] + (to + 1), sizeof(short) * (up[i][j] - (to + 1) + 1));
          memcpy(R[i & 1][j] + (to + 1), R[(i + 1) & 1][j] + (to + 1), sizeof(short) * (up[i][j] - (to + 1) + 1));
          for (int k = to + 1; k <= up[i][j]; k++) {
            sum += row[k][R[i & 1][j][k] + 1] - row[k][L[i & 1][j][k]];
          }
        }
      } else {
        if (i < n - 1 && a[i + 1][j] != 0) {
          L[i & 1][j][i] = j;
          R[i & 1][j][i] = j;
          up[i][j] = up[i + 1][j];
          memcpy(L[i & 1][j] + (i + 1), L[(i + 1) & 1][j] + (i + 1), sizeof(short) * (up[i][j] - (i + 1) + 1));
          memcpy(R[i & 1][j] + (i + 1), R[(i + 1) & 1][j] + (i + 1), sizeof(short) * (up[i][j] - (i + 1) + 1));
        } else {
          if (j < n - 1 && a[i][j + 1] != 0) {
            L[i & 1][j][i] = j;
            R[i & 1][j][i] = R[i & 1][j + 1][i];
            up[i][j] = up[i][j + 1];
            memcpy(L[i & 1][j] + (i + 1), L[i & 1][j + 1] + (i + 1), sizeof(short) * (up[i][j] - (i + 1) + 1));
            memcpy(R[i & 1][j] + (i + 1), R[i & 1][j + 1] + (i + 1), sizeof(short) * (up[i][j] - (i + 1) + 1));
          } else {
            L[i & 1][j][i] = j;
            R[i & 1][j][i] = j;
            up[i][j] = i;
          }
        }
        for (int k = i; k <= up[i][j]; k++) {
          sum += row[k][R[i & 1][j][k] + 1] - row[k][L[i & 1][j][k]];
        }
      }
      ans += a[i][j] * (sum - a[i][j]);
      put(i, j);
    }
  }
  cout << ans << '\n';
  cerr << "time = " << clock() << " ms" << "\n";
  return 0;
}