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

int N;
vector<string> v;

int sub(int start) {
    vector<int> used(N, -1);
    set<int> now;
    now.insert(start);
    int ans = 0;

    while (now.size() > 0) {
        set<int> next;
        ans++;
        for (auto i : now) used[i] = ans;
        for (auto i : now) {
            for (int j = 0; j < v[i].size(); j++) {
                if (v[i][j] == '1') {
                    if (used[j] != (ans - 1) && used[j] != -1) return -1;
                    if (used[j] == -1) {
                        next.insert(j);
                    }
                }
            }
        }
        now = next;
    }


    return ans;
}

int main() {
    cin >> N;
    v.resize(N);
    for (auto &i : v) cin >> i;

    int ans = -1;
    for (int i = 0; i < N; i++) {
        ans = max(ans, sub(i));
    }

    cout << ans << endl;

    return 0;
}
### Prompt

In Cpp, your task is to solve 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;
vector<string> v;

int sub(int start) {
    vector<int> used(N, -1);
    set<int> now;
    now.insert(start);
    int ans = 0;

    while (now.size() > 0) {
        set<int> next;
        ans++;
        for (auto i : now) used[i] = ans;
        for (auto i : now) {
            for (int j = 0; j < v[i].size(); j++) {
                if (v[i][j] == '1') {
                    if (used[j] != (ans - 1) && used[j] != -1) return -1;
                    if (used[j] == -1) {
                        next.insert(j);
                    }
                }
            }
        }
        now = next;
    }


    return ans;
}

int main() {
    cin >> N;
    v.resize(N);
    for (auto &i : v) cin >> i;

    int ans = -1;
    for (int i = 0; i < N; i++) {
        ans = max(ans, sub(i));
    }

    cout << ans << endl;

    return 0;
}
```