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

#define N 222

int n, d[N];
vector <int> adj[N];

int solve(int x) {
    for (int i = 1; i <= n; i ++) d[i] = 0;
    d[x] = 1;
    queue <int> Q; Q.push(x);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (auto v : adj[u]) {
            if (!d[v]) d[v] = d[u] + 1, Q.push(v);
            else if (d[v] == d[u]) return -1;
        }
    }
    int rlt = 0;
    for (int i = 1; i <= n; i ++) rlt = max(rlt, d[i]);
    return rlt;
}

char s[N];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++) {
        scanf("%s", s + 1);
        for (int j = 1; j <= n; j ++) {
            if (s[j] > '0') adj[i].push_back(j);
        }
    }
    int ans = -1;
    for (int i = 1; i <= n; i ++) ans = max(ans, solve(i));
    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
#include <bits/stdc++.h>
using namespace std;

#define N 222

int n, d[N];
vector <int> adj[N];

int solve(int x) {
    for (int i = 1; i <= n; i ++) d[i] = 0;
    d[x] = 1;
    queue <int> Q; Q.push(x);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (auto v : adj[u]) {
            if (!d[v]) d[v] = d[u] + 1, Q.push(v);
            else if (d[v] == d[u]) return -1;
        }
    }
    int rlt = 0;
    for (int i = 1; i <= n; i ++) rlt = max(rlt, d[i]);
    return rlt;
}

char s[N];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++) {
        scanf("%s", s + 1);
        for (int j = 1; j <= n; j ++) {
            if (s[j] > '0') adj[i].push_back(j);
        }
    }
    int ans = -1;
    for (int i = 1; i <= n; i ++) ans = max(ans, solve(i));
    printf("%d\n", ans);
}

```