#include <bits/stdc++.h>

using namespace std;

const int N = 222;

char s[N][N];
int n, dist[N];
bool done[N];

void bfs(int v) {
  memset(dist, -1, sizeof dist);
  dist[v] = 0;
  queue<int> que;
  que.push(v);
  while (!que.empty()) {
    v = que.front(); que.pop();
    for (int j = 0; j < n; ++j) {
      if (s[v][j] == '0') continue;
      if (dist[j] != -1) continue;
      dist[j] = dist[v] + 1;
      que.push(j);
    }
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    scanf("%s", s[i]);
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    bfs(i);
    bool ok = 1;
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        if (s[j][k] == '0') continue;
        ok &= abs(dist[j] - dist[k]) == 1;
      }
    }
    if (ok)
      ans = max(ans, 1 + * max_element(dist, dist+n));
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

const int N = 222;

char s[N][N];
int n, dist[N];
bool done[N];

void bfs(int v) {
  memset(dist, -1, sizeof dist);
  dist[v] = 0;
  queue<int> que;
  que.push(v);
  while (!que.empty()) {
    v = que.front(); que.pop();
    for (int j = 0; j < n; ++j) {
      if (s[v][j] == '0') continue;
      if (dist[j] != -1) continue;
      dist[j] = dist[v] + 1;
      que.push(j);
    }
  }
}

int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    scanf("%s", s[i]);
  int ans = -1;
  for (int i = 0; i < n; ++i) {
    bfs(i);
    bool ok = 1;
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        if (s[j][k] == '0') continue;
        ok &= abs(dist[j] - dist[k]) == 1;
      }
    }
    if (ok)
      ans = max(ans, 1 + * max_element(dist, dist+n));
  }
  printf("%d\n", ans);
  return 0;
}

```