#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

const int N = 205;

int n;
int f[N];
char G[N][N];

int bfs(int rt) {
	queue<int> Q;
	memset(f, 0, sizeof(f));
	f[rt] = 1;
	Q.push(rt);
	int ret = 1;
	while (!Q.empty()) {
		int p = Q.front();
		Q.pop();
		ret = max(ret, f[p]);
		for (int e = 1; e <= n; e++) {
			if (G[p][e] == '1') {
				if (f[e]) {
					if (f[e] == f[p])
						return -1;
				} else {
					f[e] = f[p] + 1;
					Q.push(e);
				}
			}
		}
	}
	return ret;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%s", G[i] + 1);
	}
	int ans = bfs(1);
	if (ans == -1) {
		puts("-1");
		return 0;
	}
	for (int i = 2; i <= n; i++) {
		ans = max(ans, bfs(i));
	}
	printf("%d\n", ans);
	return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

Given is a connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the edges are described by a grid of characters S. If S_{i,j} is `1`, there is an edge connecting Vertex i and j; otherwise, there is no such edge.

Determine whether it is possible to divide the vertices into non-empty sets V_1, \dots, V_k such that the following condition is satisfied. If the answer is yes, find the maximum possible number of sets, k, in such a division.

* Every edge connects two vertices belonging to two "adjacent" sets. More formally, for every edge (i,j), there exists 1\leq t\leq k-1 such that i\in V_t,j\in V_{t+1} or i\in V_{t+1},j\in V_t holds.

Constraints

* 2 \leq N \leq 200
* S_{i,j} is `0` or `1`.
* S_{i,i} is `0`.
* S_{i,j}=S_{j,i}
* The given graph is connected.
* N is an integer.

Input

Input is given from Standard Input in the following format:


N
S_{1,1}...S_{1,N}
:
S_{N,1}...S_{N,N}


Output

If it is impossible to divide the vertices into sets so that the condition is satisfied, print -1. Otherwise, print the maximum possible number of sets, k, in a division that satisfies the condition.

Examples

Input

2
01
10


Output

2


Input

3
011
101
110


Output

-1


Input

6
010110
101001
010100
101000
100000
010000


Output

4

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

const int N = 205;

int n;
int f[N];
char G[N][N];

int bfs(int rt) {
	queue<int> Q;
	memset(f, 0, sizeof(f));
	f[rt] = 1;
	Q.push(rt);
	int ret = 1;
	while (!Q.empty()) {
		int p = Q.front();
		Q.pop();
		ret = max(ret, f[p]);
		for (int e = 1; e <= n; e++) {
			if (G[p][e] == '1') {
				if (f[e]) {
					if (f[e] == f[p])
						return -1;
				} else {
					f[e] = f[p] + 1;
					Q.push(e);
				}
			}
		}
	}
	return ret;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%s", G[i] + 1);
	}
	int ans = bfs(1);
	if (ans == -1) {
		puts("-1");
		return 0;
	}
	for (int i = 2; i <= n; i++) {
		ans = max(ans, bfs(i));
	}
	printf("%d\n", ans);
	return 0;
}

```