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

typedef long long ll;
const ll MOD = 1e9 + 7;
int G[200][200];
int dp[200][200];
int color[200];
int N;
bool check(int v, int c) {
	color[v] = c;
	for (int to = 0; to < N; to++) {
		if (!G[v][to]) continue;
		if (color[to] == c) return 0;
		if (color[to] == 0 && !check(to, -c)) return 0;
	}
	return 1;
}
int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		string S;
		cin >> S;
		for (int j = 0; j < N; j++) G[i][j] = S[j] - '0';
	}
	if (!check(0, 1)) {
		cout << -1 << endl;
		return 0;
	}
	fill((int *)dp, (int *)(dp + N), 1<<25);
	for (int i = 0; i < N; i++) {
		dp[i][i] = 0;
		for (int j = 0; j < N; j++) if (G[i][j]) dp[i][j] = 1;
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
			}
		}
	}
	int d = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			d = max(d, dp[i][j]);
		}
	}
	cout << d + 1 << endl;
}
### Prompt

In Cpp, your task is to solve 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 long long ll;
const ll MOD = 1e9 + 7;
int G[200][200];
int dp[200][200];
int color[200];
int N;
bool check(int v, int c) {
	color[v] = c;
	for (int to = 0; to < N; to++) {
		if (!G[v][to]) continue;
		if (color[to] == c) return 0;
		if (color[to] == 0 && !check(to, -c)) return 0;
	}
	return 1;
}
int main() {
	cin >> N;
	for (int i = 0; i < N; i++) {
		string S;
		cin >> S;
		for (int j = 0; j < N; j++) G[i][j] = S[j] - '0';
	}
	if (!check(0, 1)) {
		cout << -1 << endl;
		return 0;
	}
	fill((int *)dp, (int *)(dp + N), 1<<25);
	for (int i = 0; i < N; i++) {
		dp[i][i] = 0;
		for (int j = 0; j < N; j++) if (G[i][j]) dp[i][j] = 1;
	}
	for (int k = 0; k < N; k++) {
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
			}
		}
	}
	int d = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			d = max(d, dp[i][j]);
		}
	}
	cout << d + 1 << endl;
}
```