#include <bits/stdc++.h>
using namespace std;
const int N = 3e2+5, oo = 0x3f3f3f3f;
int n, m, c[N], ans, d[N][N]; char s[N][N];
void dfs(int x, int w)
{
    c[x] = w;
    for(int y = 1; y <= n; ++y) if(s[x][y] == '1')
    {
        if(!c[y]) dfs(y, w^1);
        else if(c[y] != (w^1)) puts("-1"), exit(0);
    }
}
int main()
{
    scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%s", s[i]+1);
    dfs(1, 2); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) 
        if(i == j) d[i][j] = 0; else if(s[i][j] == '1') d[i][j] = 1; else d[i][j] = oo;
    for(int k = 1; k <= n; ++k) for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j)
        d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
    for(int i = 1; i <= n; ++i) ans = max(ans, *max_element(d[i]+1, d[i]+n+1));
    printf("%d\n", ans+1);
    return 0;
}
### Prompt

Please formulate 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>
using namespace std;
const int N = 3e2+5, oo = 0x3f3f3f3f;
int n, m, c[N], ans, d[N][N]; char s[N][N];
void dfs(int x, int w)
{
    c[x] = w;
    for(int y = 1; y <= n; ++y) if(s[x][y] == '1')
    {
        if(!c[y]) dfs(y, w^1);
        else if(c[y] != (w^1)) puts("-1"), exit(0);
    }
}
int main()
{
    scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%s", s[i]+1);
    dfs(1, 2); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) 
        if(i == j) d[i][j] = 0; else if(s[i][j] == '1') d[i][j] = 1; else d[i][j] = oo;
    for(int k = 1; k <= n; ++k) for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j)
        d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
    for(int i = 1; i <= n; ++i) ans = max(ans, *max_element(d[i]+1, d[i]+n+1));
    printf("%d\n", ans+1);
    return 0;
}
```