#include <bits/stdc++.h>

using namespace std;
//首发题解+年后复苏第一题，差点崩了！！！！！
int n, k;
bool vis[3001][3001];
char a[3001][3001];
struct node {
	int x, y, t;//x,y记录坐标，t记录时间
};
const int dx[8] = {0, 0, 1, -1, -1, 1, -1, 1};
const int dy[8] = {1, -1, 0, 0, -1, 1, 1, -1};
//int nck[3]={2,1,0};这个*数组硬控30分钟阳寿
//废话不多说，开始正片
int bfs(int x, int y) {
	//标准BFS开头
	queue<node> q;
	q.push({x, y, 0});
	vis[x][y] = 1;
	while (!q.empty()) {
		node fg = q.front();
		int cx = fg.x;
		int cy = fg.y;
		q.pop();
		if (cx == n - 2 && cy == n - 2)
			return fg.t;
		q.push({cx, cy, fg.t + 1});//重点！站着不动也加进去(此处硬控10分钟阳寿)
		for (int i = 0; i < 4; i++) {//正常走地图
			int nx = dx[i] + cx;
			int ny = dy[i] + cy;
			int fas;
			//fas=nck[(fg.t)/k];万恶之源
			//以自身为中心
			//5*5范围
			//0 0 0 0 0
			//0 0 0 0 0
			//0 0 1 0 0
			//0 0 0 0 0
			//0 0 0 0 0
			//3*3范围
			//0 0 0
			//0 1 0
			//0 0 0
			if (fg.t < k)
				fas = 2;
			else if (fg.t < 2 * k)
				fas = 1;
			else
				fas = 0;
			//判断是否合法
			if (nx + fas > n || nx - fas < 1 || ny + fas > n || ny - fas < 1 || vis[nx][ny] == 1) {
				continue;
			}
			bool fld = true;
			//判断以自身为中心的5*5或3*3或1*1范围内是否有障碍物
			if (fas != 0) {
				for (int j = nx - fas; j <= nx + fas; j++) {
					for (int k = ny - fas; k <= ny + fas; k++) {
						if (a[j][k] != '+') {
							fld = false;
						}
					}
				}
			} else {
				if (a[nx][ny] != '+') {
					fld = false;
				}
			}
			if (fld != true)
				continue;
			//加入队列
			vis[nx][ny] = 1;
			q.push({nx, ny, fg.t + 1});
		}
	}
	return -1;
}
int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}
	cout << bfs(3, 3);

	return 0;
}