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

int e[200][200];

int calc(int f, int n) {
    vector<int> c(n, -1);
    queue<int> q; q.push(f); c[f] = 1; int ret = 1;
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        for (int i = 0; i < n; i++) {
            if (e[curr][i]) {
                if (c[i] == -1) {
                    c[i] = c[curr] + 1;
                    ret = max(ret, c[i]);
                    q.push(i);
                } else {
                    if (abs(c[i] - c[curr]) != 1) {
                        return -1;
                    }
                }
            }
        }
    }
    return ret;
}

int main() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        string s; cin >> s;
        for (int j = 0; j < n; j++) {
            e[i][j] = s[j] - '0';
        }
    }
    int ans = -1;
    for (int i = 0; i < n; i++) {
        ans = max(ans, calc(i, n));
    }
    cout << ans << endl;
    return 0;
}
### Prompt

Your task is to create 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 e[200][200];

int calc(int f, int n) {
    vector<int> c(n, -1);
    queue<int> q; q.push(f); c[f] = 1; int ret = 1;
    while (!q.empty()) {
        int curr = q.front(); q.pop();
        for (int i = 0; i < n; i++) {
            if (e[curr][i]) {
                if (c[i] == -1) {
                    c[i] = c[curr] + 1;
                    ret = max(ret, c[i]);
                    q.push(i);
                } else {
                    if (abs(c[i] - c[curr]) != 1) {
                        return -1;
                    }
                }
            }
        }
    }
    return ret;
}

int main() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        string s; cin >> s;
        for (int j = 0; j < n; j++) {
            e[i][j] = s[j] - '0';
        }
    }
    int ans = -1;
    for (int i = 0; i < n; i++) {
        ans = max(ans, calc(i, n));
    }
    cout << ans << endl;
    return 0;
}
```