#include <bits/stdc++.h>
#define INF 1e18
#define int long long
#define Rep(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) Rep(i, 0, n)
#define all(a) (a).begin(), (a).end()
using namespace std;
typedef pair<int, int> P;
typedef pair<int, P> PP;
const int mod = 1000000007;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;
    int d[210][210];
    rep(i, n) rep(j, n) d[i][j] = INF;
    rep(i, n) d[i][i] = 0;
    rep(i, n)
    {
        string s;
        cin >> s;
        rep(j, n) if (s[j] == '1') d[i][j] = 1;
    }
    rep(k, n) rep(i, n) rep(j, n) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    rep(i, n) rep(j, n) rep(k, n)
    {
        int len = d[i][j] + d[j][k] + d[k][i];
        if (len % 2 == 1)
        {
            cout << -1 << endl;
            return 0;
        }
    }
    int ans = -1;
    rep(i, n) rep(j, n) ans = max(ans, d[i][j] + 1);
    cout << ans << endl;
}
### 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>
#define INF 1e18
#define int long long
#define Rep(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) Rep(i, 0, n)
#define all(a) (a).begin(), (a).end()
using namespace std;
typedef pair<int, int> P;
typedef pair<int, P> PP;
const int mod = 1000000007;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n;
    cin >> n;
    int d[210][210];
    rep(i, n) rep(j, n) d[i][j] = INF;
    rep(i, n) d[i][i] = 0;
    rep(i, n)
    {
        string s;
        cin >> s;
        rep(j, n) if (s[j] == '1') d[i][j] = 1;
    }
    rep(k, n) rep(i, n) rep(j, n) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    rep(i, n) rep(j, n) rep(k, n)
    {
        int len = d[i][j] + d[j][k] + d[k][i];
        if (len % 2 == 1)
        {
            cout << -1 << endl;
            return 0;
        }
    }
    int ans = -1;
    rep(i, n) rep(j, n) ans = max(ans, d[i][j] + 1);
    cout << ans << endl;
}
```