#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int n;
vector< string > s;
vector< int > grp;
int f(int v){
    int res = 0;
    queue<int> q;
    q.push(v);
    grp[v] = 1;
    while(!q.empty()){
        int x = q.front(); q.pop();
        if(res < grp[x]) res = grp[x];
        for(int i=0;i<n;i++){
            if(s[x][i] == '1'){
                if(grp[i] == -1){
                    q.push(i);
                    grp[i] = grp[x] + 1;
                }else if(grp[i] != grp[x]-1 && grp[i] != grp[x]+1){
                    return -1;
                }
            }
        }
    }
    return res;
}
int main(void){
    cin >> n;
    s.assign(n, "");
    for(int i=0;i<n;i++) cin >> s[i];
    int ans = -1;
    for(int i=0;i<n;i++){
        grp.assign(n, -1);
        int tmp = f(i);
        if(ans < tmp) ans = tmp;
    }
    cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <iostream>
#include <vector>
#include <queue>
using namespace std;
int n;
vector< string > s;
vector< int > grp;
int f(int v){
    int res = 0;
    queue<int> q;
    q.push(v);
    grp[v] = 1;
    while(!q.empty()){
        int x = q.front(); q.pop();
        if(res < grp[x]) res = grp[x];
        for(int i=0;i<n;i++){
            if(s[x][i] == '1'){
                if(grp[i] == -1){
                    q.push(i);
                    grp[i] = grp[x] + 1;
                }else if(grp[i] != grp[x]-1 && grp[i] != grp[x]+1){
                    return -1;
                }
            }
        }
    }
    return res;
}
int main(void){
    cin >> n;
    s.assign(n, "");
    for(int i=0;i<n;i++) cin >> s[i];
    int ans = -1;
    for(int i=0;i<n;i++){
        grp.assign(n, -1);
        int tmp = f(i);
        if(ans < tmp) ans = tmp;
    }
    cout << ans << endl;
}

```