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

int col[205], dist[205][205];
vector<int> G[205];

bool dfs(int u, int d) {
	col[u] = d % 2;
	for (auto& v : G[u]) {
		if (col[u] == col[v] || (col[v] == -1 && !dfs(v, d + 1))) return false;
	}
	return true;
}

int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) dist[i][j] = (i != j) * 1e9;
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			char c;
			cin >> c;
			if (c == '1') {
				G[i].push_back(j);
				dist[i][j] = 1;
			}
		}
	}

	memset(col, -1, 4 * N);
	if (!dfs(0, 0)) {
		cout << -1 << '\n';
		return 0;
	}	
	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][k] + dist[k][j], dist[i][j]);
		}
	}

	int ans = 0;
	for (int i = 0; i < N; ++i) ans = max(*max_element(dist[i], dist[i] + N), ans);
	cout << ans + 1 << '\n';
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 col[205], dist[205][205];
vector<int> G[205];

bool dfs(int u, int d) {
	col[u] = d % 2;
	for (auto& v : G[u]) {
		if (col[u] == col[v] || (col[v] == -1 && !dfs(v, d + 1))) return false;
	}
	return true;
}

int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) dist[i][j] = (i != j) * 1e9;
	}
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			char c;
			cin >> c;
			if (c == '1') {
				G[i].push_back(j);
				dist[i][j] = 1;
			}
		}
	}

	memset(col, -1, 4 * N);
	if (!dfs(0, 0)) {
		cout << -1 << '\n';
		return 0;
	}	
	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][k] + dist[k][j], dist[i][j]);
		}
	}

	int ans = 0;
	for (int i = 0; i < N; ++i) ans = max(*max_element(dist[i], dist[i] + N), ans);
	cout << ans + 1 << '\n';
	return 0;
}
```