#include <bits/stdc++.h>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int INF=1001001001;
vector<int>to[205];
int dp[205][205];
int color[205];

bool dfs(int v, int c=1){
    color[v]=c;
    for(int u:to[v]){
        if(color[u]==c)return false;
        if(color[u]==0&&!dfs(u,-c))return false;
    }
    return true;
}

int main(){
    int n;
    cin>>n;
    rep(i,n)rep(j,n){
        char c; cin>>c;
        if(c=='1')to[i].push_back(j);
    }
    //二部グラフ判定
    if(!dfs(0)){
        cout<<-1<<endl;
        return 0;
    }
    //WF
    rep(i,205)rep(j,205){
        dp[i][j]=INF;
        if(i==j)dp[i][j]=0;
    }
    rep(i,n)rep(j,to[i].size()){
        dp[i][to[i][j]]=1;
    }
    rep(k,n)rep(i,n)rep(j,n){
        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
    }
    int ans=0;
    rep(i,n)rep(j,n){
        ans=max(ans,dp[i][j]);
    }
    ans++;
    cout<<ans<<endl;
}
### Prompt

Your task is to create 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>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int INF=1001001001;
vector<int>to[205];
int dp[205][205];
int color[205];

bool dfs(int v, int c=1){
    color[v]=c;
    for(int u:to[v]){
        if(color[u]==c)return false;
        if(color[u]==0&&!dfs(u,-c))return false;
    }
    return true;
}

int main(){
    int n;
    cin>>n;
    rep(i,n)rep(j,n){
        char c; cin>>c;
        if(c=='1')to[i].push_back(j);
    }
    //二部グラフ判定
    if(!dfs(0)){
        cout<<-1<<endl;
        return 0;
    }
    //WF
    rep(i,205)rep(j,205){
        dp[i][j]=INF;
        if(i==j)dp[i][j]=0;
    }
    rep(i,n)rep(j,to[i].size()){
        dp[i][to[i][j]]=1;
    }
    rep(k,n)rep(i,n)rep(j,n){
        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
    }
    int ans=0;
    rep(i,n)rep(j,n){
        ans=max(ans,dp[i][j]);
    }
    ans++;
    cout<<ans<<endl;
}
```