#include <bits/stdc++.h>
using namespace std;

int n, c, d[210][210], v[210];

int main() {
    cin >> n;
    for(int i = 0; i < n; i++) {
        getchar();
        for(int j = 0; j < n; j++) d[i][j] = getchar() - '0';
    }
    int ans = -1;
    for(int i = 0; i < n; i++) {
        queue<int> q;
        q.push(i);
        memset(v, 0, sizeof(v));
        v[i] = 1;
        int f = 0, tans = 0;
        while(!q.empty()) {
            int a = q.front();q.pop();
            tans = max(tans, v[a]);
            for(int j = 0; j < n; j++) {
                if(d[a][j]) {
                    if(!v[j]) v[j] = v[a] + 1, q.push(j);
                    else if(abs(v[a] - v[j]) != 1) f = 1;
                }
            }
        }
        if(!f) ans = max(ans, tans);
    }
    cout << ans;
}
### 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;

int n, c, d[210][210], v[210];

int main() {
    cin >> n;
    for(int i = 0; i < n; i++) {
        getchar();
        for(int j = 0; j < n; j++) d[i][j] = getchar() - '0';
    }
    int ans = -1;
    for(int i = 0; i < n; i++) {
        queue<int> q;
        q.push(i);
        memset(v, 0, sizeof(v));
        v[i] = 1;
        int f = 0, tans = 0;
        while(!q.empty()) {
            int a = q.front();q.pop();
            tans = max(tans, v[a]);
            for(int j = 0; j < n; j++) {
                if(d[a][j]) {
                    if(!v[j]) v[j] = v[a] + 1, q.push(j);
                    else if(abs(v[a] - v[j]) != 1) f = 1;
                }
            }
        }
        if(!f) ans = max(ans, tans);
    }
    cout << ans;
}
```