#include <bits/stdc++.h>

const int MAX_N = 210;

void fail() {
	puts("-1");
	exit(0);
}

int n;
int a[MAX_N][MAX_N], dis[MAX_N], que[MAX_N];
bool vis[MAX_N], col[MAX_N];

void dfs(int x) {
	vis[x] = true;
	for (int i = 1; i <= n; ++i) {
		if (a[x][i]) {
			if (!vis[i]) {
				col[i] = col[x] ^ 1;
				dfs(i);
			} else if (col[x] == col[i]) fail();
		}
	}
}

void bfs(int s) {
	int ql = 1, qr = 1;
	memset(dis, 0, (n + 1) * sizeof(dis[0]));
	que[1] = s;
	dis[s] = 1;
	while (ql <= qr) {
		int x = que[ql++];
		for (int i = 1; i <= n; ++i) {
			if (a[x][i] && !dis[i]) {
				dis[i] = dis[x] + 1;
				que[++qr] = i;
			}
		}
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			char c = getchar();
			while (c != '0' && c != '1') c = getchar();
			a[i][j] = c == '1';
		}
	}

	dfs(1);
	
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		bfs(i);
		for (int j = 1; j <= n; ++j) {
			ans = std::max(ans, dis[j]);
		}
	}
	printf("%d\n", ans);
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>

const int MAX_N = 210;

void fail() {
	puts("-1");
	exit(0);
}

int n;
int a[MAX_N][MAX_N], dis[MAX_N], que[MAX_N];
bool vis[MAX_N], col[MAX_N];

void dfs(int x) {
	vis[x] = true;
	for (int i = 1; i <= n; ++i) {
		if (a[x][i]) {
			if (!vis[i]) {
				col[i] = col[x] ^ 1;
				dfs(i);
			} else if (col[x] == col[i]) fail();
		}
	}
}

void bfs(int s) {
	int ql = 1, qr = 1;
	memset(dis, 0, (n + 1) * sizeof(dis[0]));
	que[1] = s;
	dis[s] = 1;
	while (ql <= qr) {
		int x = que[ql++];
		for (int i = 1; i <= n; ++i) {
			if (a[x][i] && !dis[i]) {
				dis[i] = dis[x] + 1;
				que[++qr] = i;
			}
		}
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			char c = getchar();
			while (c != '0' && c != '1') c = getchar();
			a[i][j] = c == '1';
		}
	}

	dfs(1);
	
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		bfs(i);
		for (int j = 1; j <= n; ++j) {
			ans = std::max(ans, dis[j]);
		}
	}
	printf("%d\n", ans);
}
```