#include<bits/stdc++.h>
using namespace std;
char s[300][300];
int nm[300];
int n;
int d[300][300];
int main() {
    scanf("%d",&n);
    for(int i=0;i<n;i++) {
        scanf("%s",s[i]);
        for(int j=0;j<n;j++) {
            if(s[i][j]=='1') d[i][j]=1;
            else d[i][j]=n;
        }
        d[i][i]=0;
    }
    for(int k=0;k<n;k++) {
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
            }
        }
    }
    int u=0;
    int v=0;
    for(int i=0;i<n;i++) {
        for(int j=0;j<n;j++) {
            if(d[i][j]>d[u][v]) {
                u=i;
                v=j;
            }
        }
    }
    for(int i=0;i<n;i++) nm[i]=d[u][i]+1;
    for(int i=0;i<n;i++) {
        for(int j=0;j<n;j++) {
            if(s[i][j]=='1' && abs(nm[i]-nm[j])!=1) {
                printf("-1");
                return 0;
            }
        }
    }
    printf("%d",d[u][v]+1);
    return 0;
}

### 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>
using namespace std;
char s[300][300];
int nm[300];
int n;
int d[300][300];
int main() {
    scanf("%d",&n);
    for(int i=0;i<n;i++) {
        scanf("%s",s[i]);
        for(int j=0;j<n;j++) {
            if(s[i][j]=='1') d[i][j]=1;
            else d[i][j]=n;
        }
        d[i][i]=0;
    }
    for(int k=0;k<n;k++) {
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
            }
        }
    }
    int u=0;
    int v=0;
    for(int i=0;i<n;i++) {
        for(int j=0;j<n;j++) {
            if(d[i][j]>d[u][v]) {
                u=i;
                v=j;
            }
        }
    }
    for(int i=0;i<n;i++) nm[i]=d[u][i]+1;
    for(int i=0;i<n;i++) {
        for(int j=0;j<n;j++) {
            if(s[i][j]=='1' && abs(nm[i]-nm[j])!=1) {
                printf("-1");
                return 0;
            }
        }
    }
    printf("%d",d[u][v]+1);
    return 0;
}

```