Problem: Note the unusual memory limit.

For a rectangular grid where each square is painted white or black, we define its complexity as follows:

* If all the squares are black or all the squares are white, the complexity is 0.
* Otherwise, divide the grid into two subgrids by a line parallel to one of the sides of the grid, and let c_1 and c_2 be the complexities of the subgrids. There can be multiple ways to perform the division, and let m be the minimum value of \max(c_1, c_2) in those divisions. The complexity of the grid is m+1.



You are given a grid with H horizontal rows and W vertical columns where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.

Find the complexity of the given grid.

Constraints

* 1 \leq H,W \leq 185
* A_{ij} is `#` or `.`.

Input

Input is given from Standard Input in the following format:


H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}


Output

Print the complexity of the given grid.

Examples

Input

3 3
...
.##
.##


Output

2


Input

6 7
.####.#
....#.
....#.
....#.
.####.#
....##


Output

4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

vector<string> a;

const int MAXN = 190;

uint8_t xb_max[MAXN][MAXN][MAXN][2];
uint8_t yb_max[MAXN][MAXN][MAXN][2];

int main(){
	int n, m;
	cin >> n >> m;
	a.resize(n);
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int _q = 0; ; _q++){
		int q = _q & 1;
		for(int xa = n; xa >= 0; xa--){
			for(int ya = m; ya >= 0; ya--){
				for(int xb = xa; xb <= n; xb++){
					if(_q == 0){
						if(ya == m || xb == xa){
							yb_max[xa][ya][xb][q] = m;
						} else if(xb == xa+1){
							yb_max[xa][ya][xb][q] = yb_max[xa][ya+1][xb][q];
							if(ya + 1 < m && a[xa][ya] != a[xa][ya+1]){
								yb_max[xa][ya][xb][q] = ya + 1;
							}
						} else {
							yb_max[xa][ya][xb][q] = min(
								yb_max[xa][ya][xb-1][q],
								yb_max[xa+1][ya][xb][q]);
							if(a[xa][ya] != a[xa+1][ya]){
								yb_max[xa][ya][xb][q] = ya;
							}
						}
					} else {
						int v = yb_max[xa][ya][xb][q^1];
						yb_max[xa][ya][xb][q] = yb_max[xa][v][xb][q^1];
					}
				}
				for(int yb = ya; yb <= m; yb++){
					if(_q == 0){
						if(xa == n || yb == ya){
							xb_max[xa][ya][yb][q] = m;
						} else if(yb == ya + 1){
							xb_max[xa][ya][yb][q] = xb_max[xa+1][ya][yb][q];
							if(xa + 1 < n && a[xa][ya] != a[xa+1][ya]){
								xb_max[xa][ya][yb][q] = xa + 1;
							}
						} else {
							xb_max[xa][ya][yb][q] = min(
								xb_max[xa][ya][yb-1][q],
								xb_max[xa][ya+1][yb][q]);
							if(a[xa][ya] != a[xa][ya+1]){
								xb_max[xa][ya][yb][q] = xa;
							}
						}
					} else {
						int v = xb_max[xa][ya][yb][q^1];
						xb_max[xa][ya][yb][q] = xb_max[v][ya][yb][q^1];
					}
				}
				// merge here
				uint8_t cury = ya;
				for(int xb = n; xb >= xa; xb--){
					while(cury < m && xb_max[xa][ya][cury+1][q] >= xb){
						cury++;
					}
					yb_max[xa][ya][xb][q] = max(yb_max[xa][ya][xb][q], cury);
				}
				uint8_t curx = xa;
				for(int yb = m; yb >= ya; yb--){
					while(curx < n && yb_max[xa][ya][curx+1][q] >= yb){
						curx++;
					}
					xb_max[xa][ya][yb][q] = max(xb_max[xa][ya][yb][q], curx);
				}
			}
		}
		if(yb_max[0][0][n][q] == m){
			cout << _q << '\n';
			break;
		}
	}
}