#include <bits/stdc++.h>
using namespace std;
int a[210][210], col[210], n;
char s[210];
inline bool dfs(int x, int c) {
	col[x] = c;
	for(int i = 1; i <= n; ++ i) if(a[x][i] == 1) {
		if(col[i]) {
			if(col[i] + c != 3) return false;
		} else {
			if(dfs(i, 3 - c) == false) return false;
		}
	} return true;
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i) {
		scanf("%s", s + 1);
		for(int j = 1; j <= n; ++ j) {
			a[i][j] = s[j] - '0';
			if(!a[i][j] && i != j) a[i][j] = 1e9;
		}
	} if(dfs(1, 1) == false) return puts("-1"), 0;
	for(int k = 1; k <= n; ++ k) {
		for(int i = 1; i <= n; ++ i) {
			for(int j = 1; j <= n; ++ j) {
				a[i][j] = min(a[i][k] + a[k][j], a[i][j]);
			}
		}
	} int mx = 0;
	for(int i = 1; i <= n; ++ i) {
		for(int j = 1; j <= n; ++ j) {
			mx = max(mx, a[i][j]);
		}
	} printf("%d\n", mx + 1);
}
### 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;
int a[210][210], col[210], n;
char s[210];
inline bool dfs(int x, int c) {
	col[x] = c;
	for(int i = 1; i <= n; ++ i) if(a[x][i] == 1) {
		if(col[i]) {
			if(col[i] + c != 3) return false;
		} else {
			if(dfs(i, 3 - c) == false) return false;
		}
	} return true;
}
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i) {
		scanf("%s", s + 1);
		for(int j = 1; j <= n; ++ j) {
			a[i][j] = s[j] - '0';
			if(!a[i][j] && i != j) a[i][j] = 1e9;
		}
	} if(dfs(1, 1) == false) return puts("-1"), 0;
	for(int k = 1; k <= n; ++ k) {
		for(int i = 1; i <= n; ++ i) {
			for(int j = 1; j <= n; ++ j) {
				a[i][j] = min(a[i][k] + a[k][j], a[i][j]);
			}
		}
	} int mx = 0;
	for(int i = 1; i <= n; ++ i) {
		for(int j = 1; j <= n; ++ j) {
			mx = max(mx, a[i][j]);
		}
	} printf("%d\n", mx + 1);
}
```