There is a square grid of size n × n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h, w) to color a rectangle of size h × w. You are to make all cells white for minimum total cost.

Input

The first line contains a single integer n (1 ≤ n ≤ 50) — the size of the square grid.

Each of the next n lines contains a string of length n, consisting of characters '.' and '#'. The j-th character of the i-th line is '#' if the cell with coordinates (i, j) is black, otherwise it is white.

Output

Print a single integer — the minimum total cost to paint all cells in white.

Examples

Input


3
###
#.#
###


Output


3


Input


3
...
...
...


Output


0


Input


4
#...
....
....
#...


Output


2


Input


5
#...#
.#.#.
.....
.#...
#....


Output


5

Note

The examples and some of optimal solutions are shown on the pictures below.

<image>
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e17;
mt19937 random1(136786931);
int randlr(int l, int r) { return random1() % (r - l + 1) + l; }
const long double PI = 3.1415926535897932384626433832795028841971;
struct Point {
  int first, second;
  Point(){};
  Point(int _x, int _y) {
    first = _x;
    second = _y;
  }
};
struct Vector {
  int first, second;
  Vector(){};
  Vector(Point p) {
    first = p.first;
    second = p.second;
  }
  Vector(Point p1, Point p2) {
    first = p2.first - p1.first;
    second = p2.second - p1.second;
  }
  int operator*(Vector &oth) { return first * oth.second - second * oth.first; }
};
bool in_ang(Point p, Point a, Point o, Point b) {
  Vector oa(o, a), ob(o, b), op(o, p);
  return (oa * op > 0 && op * ob > 0) || (oa * op < 0 && op * ob < 0);
}
const long double eps = 1e-7;
bool eq(long double first, long double second) {
  return abs(first - second) <= eps;
}
const int mod = (1e9) + 7;
vector<vector<int>> mult_matr(vector<vector<int>> a, vector<vector<int>> b) {
  int n = a.size();
  vector<vector<int>> c(n);
  for (int i = 0; i < n; i++) {
    c[i].resize(b[0].size());
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < b.size(); j++) {
      for (int k = 0; k < b.size(); k++) {
        c[i][j] += a[i][k] * b[k][j];
        c[i][j] %= (mod - 1);
      }
    }
  }
  return c;
}
vector<vector<int>> fpow(vector<vector<int>> a, int b) {
  vector<vector<int>> answ = a;
  b--;
  while (b) {
    if (b & 1) {
      answ = mult_matr(answ, a);
    }
    a = mult_matr(a, a);
    b >>= 1;
  }
  return answ;
}
int fpow(int first, int second) {
  int answ = 1;
  while (second) {
    if (second & 1) {
      answ *= first;
      answ %= mod;
    }
    first *= first;
    first %= mod;
    second >>= 1;
  }
  return answ;
}
int del(int a, int b) { return (a * fpow(b, mod - 2)) % mod; }
char dp[51][51][51][51];
bool field[51][51];
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      char c;
      cin >> c;
      if (c == '#') field[i][j] = 1;
      dp[i][j][i][j] = field[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int i1 = i; i1 < n; i1++) {
        for (int j1 = j + (i1 == i); j1 < n; j1++) {
          dp[i][j][i1][j1] = max(j1 - j + 1, i1 - i + 1);
        }
      }
    }
  }
  for (int leni = 0; leni < n; leni++) {
    for (int lenj = 0; lenj < n; lenj++) {
      for (int i = 0; i < n - leni; i++) {
        for (int j = 0; j < n - lenj; j++) {
          int i1 = i + leni, j1 = j + lenj;
          for (int k = i; k < i1; k++) {
            dp[i][j][i1][j1] = min((int)dp[i][j][i1][j1],
                                   dp[i][j][k][j1] + dp[k + 1][j][i1][j1]);
          }
          for (int k = j; k < j1; k++) {
            dp[i][j][i1][j1] = min((int)dp[i][j][i1][j1],
                                   dp[i][j][i1][k] + dp[i][k + 1][i1][j1]);
          }
        }
      }
    }
  }
  cout << (int)dp[0][0][n - 1][n - 1];
  return 0;
}
