#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int N = 200;
const int INF = 1e9;

int n;
char s[N][N];

int dst[N][N];
bool visited[N];

void dfs(int pos, bool *bg) {
  visited[pos] = true;
  rep(to, n) {
    if (s[pos][to] == '0') continue;
    if (pos == to) continue;
    if (dst[0][pos] == dst[0][to]) *bg = false;
    if (visited[to]) continue;
    dfs(to, bg);
  }
}

int main() {
  cin >> n;
  rep(i, n) rep(j, n) cin >> s[i][j];

  rep(i, n) rep(j, n) {
    dst[i][j] = INF;
    if (i == j) dst[i][j] = 0;
    if (s[i][j] == '1') dst[i][j] = 1;
  }

  rep(k, n) rep(i, n) rep(j, n) dst[i][j] = min(dst[i][j], dst[i][k] + dst[k][j]);

  rep(i, n) visited[i] = false;
  bool bg = true;
  dfs(0, &bg);

  if (!bg) {
    cout << -1 << endl;
    return 0;
  }

  int ans = 0;
  rep(i, n) rep(j, n) ans = max(ans, dst[i][j] + 1);
  cout << ans << endl;
  return 0;
}
### Prompt

Please create a solution in Cpp 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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const int N = 200;
const int INF = 1e9;

int n;
char s[N][N];

int dst[N][N];
bool visited[N];

void dfs(int pos, bool *bg) {
  visited[pos] = true;
  rep(to, n) {
    if (s[pos][to] == '0') continue;
    if (pos == to) continue;
    if (dst[0][pos] == dst[0][to]) *bg = false;
    if (visited[to]) continue;
    dfs(to, bg);
  }
}

int main() {
  cin >> n;
  rep(i, n) rep(j, n) cin >> s[i][j];

  rep(i, n) rep(j, n) {
    dst[i][j] = INF;
    if (i == j) dst[i][j] = 0;
    if (s[i][j] == '1') dst[i][j] = 1;
  }

  rep(k, n) rep(i, n) rep(j, n) dst[i][j] = min(dst[i][j], dst[i][k] + dst[k][j]);

  rep(i, n) visited[i] = false;
  bool bg = true;
  dfs(0, &bg);

  if (!bg) {
    cout << -1 << endl;
    return 0;
  }

  int ans = 0;
  rep(i, n) rep(j, n) ans = max(ans, dst[i][j] + 1);
  cout << ans << endl;
  return 0;
}
```