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

const int N = 220;

int n;
char pan[N][N];
int vis[N], dp[N], que[N];

bool dfs(int u, int c) {
	vis[u] = c;
	for (int i = 1; i <= n; i++) if (pan[u][i] == '1') {
		if (vis[i] != -1 && vis[i] == c) return 0;
		if (vis[i] == -1 && !dfs(i, !c)) return 0;
	}
	return 1;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> pan[i] + 1;
	for (int i = 1; i <= n; i++) vis[i] = -1;
	if (!dfs(1, 0)) {
		cout << "-1\n";
		return 0;
	}
	int ans = 0;
	for (int s = 1; s <= n; s++) {
		for (int i = 1; i <= n; i++) dp[i] = 0;
		int hd = 0, tl = 0;
		dp[s] = 1, que[tl++] = s;
		while (hd < tl) {
			int cur = que[hd++];
			for (int i = 1; i <= n; i++) if (pan[cur][i] == '1' && !dp[i]) dp[i] = dp[cur] + 1, que[tl++] = i;
		}
		ans = max(ans, dp[que[tl - 1]]);
	}
	cout << ans << "\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 N = 220;

int n;
char pan[N][N];
int vis[N], dp[N], que[N];

bool dfs(int u, int c) {
	vis[u] = c;
	for (int i = 1; i <= n; i++) if (pan[u][i] == '1') {
		if (vis[i] != -1 && vis[i] == c) return 0;
		if (vis[i] == -1 && !dfs(i, !c)) return 0;
	}
	return 1;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> pan[i] + 1;
	for (int i = 1; i <= n; i++) vis[i] = -1;
	if (!dfs(1, 0)) {
		cout << "-1\n";
		return 0;
	}
	int ans = 0;
	for (int s = 1; s <= n; s++) {
		for (int i = 1; i <= n; i++) dp[i] = 0;
		int hd = 0, tl = 0;
		dp[s] = 1, que[tl++] = s;
		while (hd < tl) {
			int cur = que[hd++];
			for (int i = 1; i <= n; i++) if (pan[cur][i] == '1' && !dp[i]) dp[i] = dp[cur] + 1, que[tl++] = i;
		}
		ans = max(ans, dp[que[tl - 1]]);
	}
	cout << ans << "\n";
	return 0;
}
```