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

const int INF = 1e8;

const int MAXN = 210;
int N;
char adj[MAXN][MAXN];

int dist[MAXN][MAXN];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> adj[i];
	}

	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (i == j) dist[i][j] = 0;
			else if (adj[i][j] == '1') dist[i][j] = 1;
			else dist[i][j] = INF;
		}
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
			}
		}
	}

	for (int st = 0; st < N; st++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (adj[i][j] == '1' && dist[st][i] == dist[st][j]) {
					cout << -1 << '\n';
					exit(0);
				}
			}
		}
	}

	int diam = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			diam = max(diam, dist[i][j]);
		}
	}
	cout << diam + 1 << '\n';

	return 0;
}

### Prompt

Please create a solution in Cpp to 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;

const int INF = 1e8;

const int MAXN = 210;
int N;
char adj[MAXN][MAXN];

int dist[MAXN][MAXN];

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> adj[i];
	}

	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			if (i == j) dist[i][j] = 0;
			else if (adj[i][j] == '1') dist[i][j] = 1;
			else dist[i][j] = INF;
		}
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
			}
		}
	}

	for (int st = 0; st < N; st++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (adj[i][j] == '1' && dist[st][i] == dist[st][j]) {
					cout << -1 << '\n';
					exit(0);
				}
			}
		}
	}

	int diam = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			diam = max(diam, dist[i][j]);
		}
	}
	cout << diam + 1 << '\n';

	return 0;
}

```