#include<iostream>
#include<cstdio>
#include<queue>
const int N=205;
int dis[N],n;
char s[N][N];
int bfs(int u) {
    std::queue<int> q;
    for(int i=1;i<=n;i++) dis[i]=N;
    dis[u]=1;
    q.push(u);
    while(!q.empty()) {
        u=q.front();q.pop();
        for(int j=1;j<=n;j++) {
            if(u==j||s[u][j]!='1') continue;
            if(dis[j]>dis[u]+1) {
                dis[j]=dis[u]+1;
                q.push(j);
            }
        }
    }
    int res=1;
    for(int i=1;i<=n;i++) res=std::max(res,dis[i]);
    for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) {
        if(s[i][j]=='0') continue;
        if((dis[i]+1!=dis[j])&&dis[j]+1!=dis[i]) {
            res=-1;
        }
    }
    return res;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
    int ans=-1;
    for(int i=1;i<=n;i++) ans=std::max(ans,bfs(i));
    printf("%d",ans);
    return 0;
}

### Prompt

Please create a solution in Cpp 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<iostream>
#include<cstdio>
#include<queue>
const int N=205;
int dis[N],n;
char s[N][N];
int bfs(int u) {
    std::queue<int> q;
    for(int i=1;i<=n;i++) dis[i]=N;
    dis[u]=1;
    q.push(u);
    while(!q.empty()) {
        u=q.front();q.pop();
        for(int j=1;j<=n;j++) {
            if(u==j||s[u][j]!='1') continue;
            if(dis[j]>dis[u]+1) {
                dis[j]=dis[u]+1;
                q.push(j);
            }
        }
    }
    int res=1;
    for(int i=1;i<=n;i++) res=std::max(res,dis[i]);
    for(int i=1;i<n;i++) for(int j=i+1;j<=n;j++) {
        if(s[i][j]=='0') continue;
        if((dis[i]+1!=dis[j])&&dis[j]+1!=dis[i]) {
            res=-1;
        }
    }
    return res;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
    int ans=-1;
    for(int i=1;i<=n;i++) ans=std::max(ans,bfs(i));
    printf("%d",ans);
    return 0;
}

```