rows = int(input())
maze, visited, exits, path, ans = [], [], [], [], []
start_r, start_c = -1, -1

for row in range(rows):
    maze.append([])
    visited.append([])
    line = input()
    for col in range(len(line)):
        maze[row].append(line[col] == ' ')
        visited[row].append(0)
        if line[col] == 'k':
            start_r, start_c = row, col
            maze[row][col] = True
        if line[col] == ' ' and ((row == 0 or row == rows - 1) or (col == 0 or col == len(line) - 1)):
            exits.append([row, col])


# Source https://stackoverflow.com/a/60957101
def traverse(r: int, c: int):
    # if it is a wall ot it was visited, do nothing
    if not maze[r][c] or visited[r][c]:
        return
    # mark the index as visited
    visited[r][c] = 1
    # add index to the path
    path.append((r, c))
    # if the exit is reached, store the path
    # and pop th last index to traverse other paths
    if [r, c] in exits:
        # any path that reached exit are appended here
        ans.append(list(path))
        path.pop()
        return
    # loop on the neighbours. if they are valid indices, do same
    for k in range(4):
        nwr = r + [-1, 1, 0, 0][k]
        nwc = c + [0, 0, 1, -1][k]
        if 0 <= nwr < len(maze) and 0 <= nwc < len(maze[nwr]):
            traverse(nwr, nwc)
    # after finishing this index, just pop because we do not need it any more
    path.pop()
    return


traverse(start_r, start_c)

if len(ans) == 0:
    # fix for when Kate is already on the exit
    if start_c == 0 or start_c == len(maze[0]) - 1 or start_r == 0 or start_r == len(maze) - 1:
        print('Kate got out in 1 moves')
    else:
        print('Kate cannot get out')
else:
    print(f"Kate got out in {min([len(p) for p in ans])} moves")