#include <bits/stdc++.h>

using namespace std;

int n;
bool adj[201][201];
int dist[201];
int ans = 0;

bool bfs(int src) {
	queue<int> q;

	fill(dist + 1, dist + n + 1, -1);
	dist[src] = 0;
	q.push(src);

	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int v = 1; v <= n; v++) {
			if (adj[u][v]) {
				if (dist[v] == -1) {
					dist[v] = dist[u] + 1;
					q.push(v);
				} else {
					if (dist[v] != dist[u] + 1 && dist[u] != dist[v] + 1) {
						return false;
					}
				}
			}
		}
	}

	for (int i = 1; i <= n; i++) {
		ans = max(ans, dist[i] + 1);
	}

	return true;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			char ch;
			while (isspace(ch = getchar())) {
			}
			adj[i][j] = ch ^ '0';
		}
	}

	for (int i = 1; i <= n; i++) {
		if (!bfs(i)) {
			puts("-1");
			exit(0);
		}
	}
	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;

int n;
bool adj[201][201];
int dist[201];
int ans = 0;

bool bfs(int src) {
	queue<int> q;

	fill(dist + 1, dist + n + 1, -1);
	dist[src] = 0;
	q.push(src);

	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int v = 1; v <= n; v++) {
			if (adj[u][v]) {
				if (dist[v] == -1) {
					dist[v] = dist[u] + 1;
					q.push(v);
				} else {
					if (dist[v] != dist[u] + 1 && dist[u] != dist[v] + 1) {
						return false;
					}
				}
			}
		}
	}

	for (int i = 1; i <= n; i++) {
		ans = max(ans, dist[i] + 1);
	}

	return true;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			char ch;
			while (isspace(ch = getchar())) {
			}
			adj[i][j] = ch ^ '0';
		}
	}

	for (int i = 1; i <= n; i++) {
		if (!bfs(i)) {
			puts("-1");
			exit(0);
		}
	}
	printf("%d\n", ans);

	return 0;
}

```