#include <bits/stdc++.h>
#define debug(x) cerr<<#x<<" = "<<x<<'\n'

using namespace std;

int n,vis[400],tot,dis[400],ans;
vector<int> g[400];
void bfs(int i) {
    queue<int> q;
    q.push(i);
    vis[i] = ++tot, dis[i] = 0;
    while(!q.empty()) {
        i = q.front(), q.pop();
        ans = max(ans,dis[i]);
        for(int j:g[i]) {
            if(vis[j] < tot) vis[j]=tot, dis[j]=dis[i]+1, q.push(j);
            else if((dis[i]&1) == (dis[j]&1)) cout << -1 << '\n', exit(0);
        }
    }
}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    cin >> n;
    for(int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for(int j = 0; j < n; j++) if(s[j]=='1') g[i].push_back(j);
    }
    for(int i = 0; i < n; i++) bfs(i);
    cout << ans+1 << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#define debug(x) cerr<<#x<<" = "<<x<<'\n'

using namespace std;

int n,vis[400],tot,dis[400],ans;
vector<int> g[400];
void bfs(int i) {
    queue<int> q;
    q.push(i);
    vis[i] = ++tot, dis[i] = 0;
    while(!q.empty()) {
        i = q.front(), q.pop();
        ans = max(ans,dis[i]);
        for(int j:g[i]) {
            if(vis[j] < tot) vis[j]=tot, dis[j]=dis[i]+1, q.push(j);
            else if((dis[i]&1) == (dis[j]&1)) cout << -1 << '\n', exit(0);
        }
    }
}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    cin >> n;
    for(int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for(int j = 0; j < n; j++) if(s[j]=='1') g[i].push_back(j);
    }
    for(int i = 0; i < n; i++) bfs(i);
    cout << ans+1 << '\n';
}

```