#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <queue>
#define debug(...) fprintf(stderr, __VA_ARGS__)

inline int input() { int x; scanf("%d", &x); return x; }

const int maxn = 205;
char s[maxn][maxn];
int d[maxn];

int ans;
bool shit;
void bfs(int st, int n) {
	std::queue<int> q;
	d[st] = 1;
	q.push(st);

	while(!q.empty()) {
		int u = q.front();
		q.pop();
		/* debug("%d %d\n", u, d[u]); */
		if(d[u] > ans) ans = d[u];

		for(int i = 1; i <= n; i ++)
			if(s[u][i] == '1' and !d[i]) {
				d[i] = d[u] + 1;
				q.push(i);
			} else if(s[u][i] == '1' and (d[i] & 1) == (d[u] & 1))
				shit = 1;
	}
}

int main() {
	int n = input();
	for(int i = 1; i <= n; i ++)
		scanf("%s", s[i] + 1);

	for(int i = 1; i <= n; i ++) {
		/* debug("start %d\n", i); */
		for(int j = 1; j <= n; j ++)
			d[j] = 0;
		bfs(i, n);
		if(shit) {
			puts("-1");
			return 0;
		}
	}

	printf("%d\n", ans);
}

### Prompt

Your challenge is to write a Cpp solution 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
#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <queue>
#define debug(...) fprintf(stderr, __VA_ARGS__)

inline int input() { int x; scanf("%d", &x); return x; }

const int maxn = 205;
char s[maxn][maxn];
int d[maxn];

int ans;
bool shit;
void bfs(int st, int n) {
	std::queue<int> q;
	d[st] = 1;
	q.push(st);

	while(!q.empty()) {
		int u = q.front();
		q.pop();
		/* debug("%d %d\n", u, d[u]); */
		if(d[u] > ans) ans = d[u];

		for(int i = 1; i <= n; i ++)
			if(s[u][i] == '1' and !d[i]) {
				d[i] = d[u] + 1;
				q.push(i);
			} else if(s[u][i] == '1' and (d[i] & 1) == (d[u] & 1))
				shit = 1;
	}
}

int main() {
	int n = input();
	for(int i = 1; i <= n; i ++)
		scanf("%s", s[i] + 1);

	for(int i = 1; i <= n; i ++) {
		/* debug("start %d\n", i); */
		for(int j = 1; j <= n; j ++)
			d[j] = 0;
		bfs(i, n);
		if(shit) {
			puts("-1");
			return 0;
		}
	}

	printf("%d\n", ans);
}

```