//两个问题
/*
1.vis标记问题---在往下深搜一条路时  搜一个点要打上经过标记vis=1  最终回溯时要将这个表记vis变回0
2.dfs写法问题  可以用dx,dy这种写法 会更方便很多
*/
#include <iostream>
#include <cstring>
using namespace std;
int n;
int f[105][105], vis[105][105];
char s[105][105];
int si, sj, ei, ej;
int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };
void dfs(int x, int y)
{
	if (x == ei && x == ej)
		return;
	for (int i = 0; i < 4; i++)
	{
		int xx = x + dx[i];
		int yy = y + dy[i];
		if (vis[xx][yy] == 0 && (xx >= 1 && xx <= n && yy >= 1 && yy <= n && s[x][y] != s[xx][yy]))
		{
			vis[xx][yy] = 1;
			f[xx][yy] = min(f[xx][yy], f[x][y] + 1);
			dfs(xx, yy);
			vis[xx][yy] = 0;
		}
	}
}
int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			cin >> s[i][j];
			if (s[i][j] == 'A')
				si = i, sj = j;
			if (s[i][j] == 'B')
				ei = i, ej = j;
		}
	memset(f, 0x3f, sizeof(f));
	vis[si][sj] = 1;
	f[si][sj] = 0;
	dfs(si, sj);
	if (f[ei][ej] == 0x3f3f3f3f)
		cout << -1 << endl;
	else
		cout << f[ei][ej] << endl;
	return 0;
}