#include <stdio.h>
#include <queue>
typedef long long ll;

int main(void) {
  ll i, j, k, n, now, ans = 0;
  scanf("%lld", &n);
  bool flg[n], flg2[n];
  ll d[n];
  for(i = 0; i < n; ++i) flg[i] = true;
  char s[n][n + 10];
  for(i = 0; i < n; ++i) scanf("%s", s[i]);
  flg[0] = true;
  std::queue<ll> q;
  q.push(0);
  while(q.size()) {
    now = q.front(), q.pop();
    for(i = 0; i < n; ++i) if(s[now][i] == '1') {
      if(flg[i]) flg[i] = false, flg2[i] = !flg2[now], q.push(i);
      else if(flg2[i] == flg2[now]) {
        printf("-1");
        return 0;
      }
    }
  }
  for(i = 0; i < n; ++i) {
    for(j = 0; j < n; ++j) flg[j] = true;
    flg[i] = false, d[i] = 1;
    q.push(i);
    while(q.size()) {
      now = q.front(), q.pop();
      if(d[now] > ans) ans = d[now];
      for(j = 0; j < n; ++j) if(s[now][j] == '1' && flg[j]) {
        d[j] = d[now] + 1;
        q.push(j), flg[j] = false;
      }
    }
  }
  printf("%lld", ans);
  return 0;
}
### Prompt

Your challenge is to write 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 <stdio.h>
#include <queue>
typedef long long ll;

int main(void) {
  ll i, j, k, n, now, ans = 0;
  scanf("%lld", &n);
  bool flg[n], flg2[n];
  ll d[n];
  for(i = 0; i < n; ++i) flg[i] = true;
  char s[n][n + 10];
  for(i = 0; i < n; ++i) scanf("%s", s[i]);
  flg[0] = true;
  std::queue<ll> q;
  q.push(0);
  while(q.size()) {
    now = q.front(), q.pop();
    for(i = 0; i < n; ++i) if(s[now][i] == '1') {
      if(flg[i]) flg[i] = false, flg2[i] = !flg2[now], q.push(i);
      else if(flg2[i] == flg2[now]) {
        printf("-1");
        return 0;
      }
    }
  }
  for(i = 0; i < n; ++i) {
    for(j = 0; j < n; ++j) flg[j] = true;
    flg[i] = false, d[i] = 1;
    q.push(i);
    while(q.size()) {
      now = q.front(), q.pop();
      if(d[now] > ans) ans = d[now];
      for(j = 0; j < n; ++j) if(s[now][j] == '1' && flg[j]) {
        d[j] = d[now] + 1;
        q.push(j), flg[j] = false;
      }
    }
  }
  printf("%lld", ans);
  return 0;
}
```