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

#define REP(i, stop) for (signed i = 0; i < stop; ++i)

#define S(i, j) ip[i * (N + 1) + j]

const signed is = 1 << 16;
char ib[is], *ip = ib, it;

using P = pair<int, int>;
signed N;
queue<P> que;
int parity[200];
P bfs(int s) {
	REP(i, N) parity[i] = -1;
	P p;
	que.push({s, 1});
	parity[s] = 0;
	while (!que.empty()) {
		p = que.front();
		que.pop();
		int i = p.first;
		int d = p.second;
		REP(j, N) {
			if (S(i, j) == '0') continue;
			if (parity[j] == -1) {
				parity[j] = 1 - parity[i];
				que.push({j, d + 1});
			} else if (parity[i] == parity[j]) {
				puts("-1\n");
				exit(0);
			}
		}
	}
	return p;
}

signed main() {
	fread_unlocked(ib, 1, is, stdin);
	while ((it = *ip++) != '\n') {
		N = N * 10 + it - '0';
	}
	P p = bfs(1);
	p = bfs(p.first);
	printf("%d\n", p.second);
}

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

#define REP(i, stop) for (signed i = 0; i < stop; ++i)

#define S(i, j) ip[i * (N + 1) + j]

const signed is = 1 << 16;
char ib[is], *ip = ib, it;

using P = pair<int, int>;
signed N;
queue<P> que;
int parity[200];
P bfs(int s) {
	REP(i, N) parity[i] = -1;
	P p;
	que.push({s, 1});
	parity[s] = 0;
	while (!que.empty()) {
		p = que.front();
		que.pop();
		int i = p.first;
		int d = p.second;
		REP(j, N) {
			if (S(i, j) == '0') continue;
			if (parity[j] == -1) {
				parity[j] = 1 - parity[i];
				que.push({j, d + 1});
			} else if (parity[i] == parity[j]) {
				puts("-1\n");
				exit(0);
			}
		}
	}
	return p;
}

signed main() {
	fread_unlocked(ib, 1, is, stdin);
	while ((it = *ip++) != '\n') {
		N = N * 10 + it - '0';
	}
	P p = bfs(1);
	p = bfs(p.first);
	printf("%d\n", p.second);
}

```