// Create time: 2019-10-05  19:59:00
#include <bits/stdc++.h>
#define rint register int
typedef long long LL;

using namespace std;

int a[205], G[205][205];
queue <int> q;

int main() {

    int n; cin >> n;
    for (rint i = 1; i <= n; ++i)
        for (rint j = 1; j <= n; ++j) {
            char ch = getchar();
            while (ch != '0' && ch != '1') ch = getchar();
            if (ch == '1') G[i][j] = 1;
        }
    int ans = 0;
    for (rint i = 1; i <= n; ++i) {
        memset(a, 0, sizeof(a));
        a[i] = 1;
        q.push(i);
        while (!q.empty()) {
            int t = q.front(); q.pop();
            for (rint i = 1; i <= n; ++i)
                if (G[t][i]) {
                    if (!a[i]) a[i] = a[t] + 1, q.push(i);
                    else if (abs(a[i] - a[t]) != 1) return puts("-1"), 0;
                }
        }
        for (rint j = 1; j <= n; ++j)
            ans = max(ans, a[j]);
    }
    printf("%d\n", ans);

    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
// Create time: 2019-10-05  19:59:00
#include <bits/stdc++.h>
#define rint register int
typedef long long LL;

using namespace std;

int a[205], G[205][205];
queue <int> q;

int main() {

    int n; cin >> n;
    for (rint i = 1; i <= n; ++i)
        for (rint j = 1; j <= n; ++j) {
            char ch = getchar();
            while (ch != '0' && ch != '1') ch = getchar();
            if (ch == '1') G[i][j] = 1;
        }
    int ans = 0;
    for (rint i = 1; i <= n; ++i) {
        memset(a, 0, sizeof(a));
        a[i] = 1;
        q.push(i);
        while (!q.empty()) {
            int t = q.front(); q.pop();
            for (rint i = 1; i <= n; ++i)
                if (G[t][i]) {
                    if (!a[i]) a[i] = a[t] + 1, q.push(i);
                    else if (abs(a[i] - a[t]) != 1) return puts("-1"), 0;
                }
        }
        for (rint j = 1; j <= n; ++j)
            ans = max(ans, a[j]);
    }
    printf("%d\n", ans);

    return 0;
}

```