251107_7569_토마토_G5_BFS

//O(H*N*M)
//BFS 3차원 배열.

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;	

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int N, M, H;
	cin >> M>> N >> H;

	vector<vector<vector<int>>> arr;
	
	for (int i = 0; i < H; i++) {
		vector<vector<int>> deck;
		deck.assign(N, vector<int>(M, 0));
		arr.push_back(deck);
	}
	
	queue<pair<int,pair<int,int>>> q;
	for (int h = 0; h < H; h++) {
		for (int n = 0; n < N; n++) {
			for (int m = 0; m < M; m++) {
				cin >> arr[h][n][m];
				if (arr[h][n][m] == 1) {
					q.push({ h, {n, m} });
				}
			}
		}
	}

	while (!q.empty()) {
		pair<int, pair<int, int>> tom = q.front();
		q.pop();

		int h = tom.first;
		int n = tom.second.first;
		int m = tom.second.second;

		int dirn[] = { 0,0,1,-1,0,0 };
		int dirm[] = { 1,-1,0,0,0,0 };
		int dirh[] = { 0,0,0,0,1,-1 };

		for (int i = 0; i < 6; i++) {
			int forn = dirn[i] + n;
			int form = dirm[i] + m;
			int forh = dirh[i] + h;
			if (forh < 0 or forh >= H or forn < 0 or forn >= N or form < 0 or form >= M) continue;
			if (arr[forh][forn][form] <= arr[h][n][m] + 1 and arr[forh][forn][form] != 0) continue;	
			if (arr[forh][forn][form] == -1) continue;

			arr[forh][forn][form] = arr[h][n][m] + 1;
			q.push({ forh,{forn,form} });
		}
	}

	int result = 0;
	for (int h = 0; h < H; h++) {
		for (int n = 0; n < N; n++) {
			for (int m = 0; m < M; m++) {
				if (arr[h][n][m] == 0) {
					result = 0;
					goto EXIT;
				}
				result = max(result, arr[h][n][m]);
			}
		}
	}
	goto EXIT;
	EXIT:
		cout << result-1;
}