import sys
from collections import deque

M, N, H = map(int, sys.stdin.readline().split())
graph = list()

for floor in range(H):
    graph.append([])
    for _ in range(N):
        graph[floor].append(list(map(int, sys.stdin.readline().split())))

def bfs(graph, limit):
    queue = deque()
    h, n, m = limit
    max_day = 0

    dx = [1, -1, 0, 0, 0, 0]
    dy = [0, 0, 1, -1, 0, 0]
    dz = [0, 0, 0, 0, 1, -1]

    for i in range(H):
        for j in range(N):
            for k in range(M):
                if graph[i][j][k] == 1:
                    queue.append([i, j, k])

    while queue:
        z, x, y = queue.popleft()

        for i in range(6):
            nz = z + dz[i]
            nx = x + dx[i]
            ny = y + dy[i]

            if 0 <= nx and nx < n and 0 <= ny and ny < m and 0 <= nz and nz < h and graph[nz][nx][ny] == 0:
                graph[nz][nx][ny] = graph[z][x][y] + 1
                queue.append([nz, nx, ny])

    for a in range(H):
        for b in range(N):
            for c in range(M):
                # 익지 않은 토마토가 하나라도 있는 경우
                if graph[a][b][c] == 0:
                    return -1
                else:
                    max_day = max(max_day, graph[a][b][c])

    # 저장될 때부터 모든 토마토가 익어있는 상태
    if max_day == 0:
        return 0
    else:
        return max_day - 1

print(bfs(graph, [H, N, M]))

import sys
input = sys.stdin.readline

def solution():

    n, m, h = map(int, input().split())
    board = [[[*map(int, input().split())] for _ in range(m)] for _ in range(h)]
    cnt = 0
    tomatoes = []
    depth = -1

    for k in range(h):
        for i in range(m):
            for j in range(n):
                if board[k][i][j] == 0:
                    cnt += 1
                elif board[k][i][j] == 1:
                    tomatoes.append((k, i, j))

    while tomatoes:
        depth += 1
        temp = []
        for k, i, j in tomatoes:
            if k > 0 and board[k-1][i][j] == 0:
                board[k-1][i][j] = 1
                temp.append((k-1,i,j))
            if i > 0 and board[k][i-1][j] == 0:
                board[k][i-1][j] = 1
                temp.append((k,i-1,j))
            if j > 0 and board[k][i][j-1] == 0:
                board[k][i][j-1] = 1
                temp.append((k,i,j-1))
            if k < h-1 and board[k+1][i][j] == 0:
                board[k+1][i][j] = 1
                temp.append((k+1,i,j))
            if i < m-1 and board[k][i+1][j] == 0:
                board[k][i+1][j] = 1
                temp.append((k,i+1,j))
            if j < n-1 and board[k][i][j+1] == 0:
                board[k][i][j+1] = 1
                temp.append((k,i,j+1))
        cnt -= len(temp)
        tomatoes = temp

    return -1 if cnt else depth
                
print(solution())