#include <bits/stdc++.h>

using namespace std;
using ll = long long;

constexpr ll MOD = 1e9 + 7;
int mx[] = {-1, 1, 0, 0}, my[] = {0, 0, -1, 1};

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(3 * n);
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		cin >> u >> v;
		u--, v--;
		g[3 * u].push_back(3 * v + 1);
		g[3 * u + 1].push_back(3 * v + 2);
		g[3 * u + 2].push_back(3 * v);
	}
	int s, t;
	cin >> s >> t;
	s--, t--;
	queue<pair<int, int>> q;
	q.push({3 * s, 0});
	vector<int> visited(3 * n);
	while (!q.empty())
	{
		auto now = q.front();
		for (auto nxt : g[now.first])
		{
			if (!visited[nxt])
			{
				q.push({nxt, now.second + 1});
				visited[nxt] = now.second + 1;
			}
		}
		q.pop();
	}
	if (visited[3 * t])
	{
		cout << visited[3 * t] / 3 << endl;
	}
	else
	{
		cout << -1 << endl;
	}
}
### Prompt

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

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G. G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

* 2 \leq N \leq 10^5
* 0 \leq M \leq \min(10^5, N (N-1))
* 1 \leq u_i, v_i \leq N(1 \leq i \leq M)
* u_i \neq v_i (1 \leq i \leq M)
* If i \neq j, (u_i, v_i) \neq (u_j, v_j).
* 1 \leq S, T \leq N
* S \neq T

Input

Input is given from Standard Input in the following format:


N M
u_1 v_1
:
u_M v_M
S T


Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1. If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

constexpr ll MOD = 1e9 + 7;
int mx[] = {-1, 1, 0, 0}, my[] = {0, 0, -1, 1};

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(3 * n);
	for (int i = 0; i < m; ++i)
	{
		int u, v;
		cin >> u >> v;
		u--, v--;
		g[3 * u].push_back(3 * v + 1);
		g[3 * u + 1].push_back(3 * v + 2);
		g[3 * u + 2].push_back(3 * v);
	}
	int s, t;
	cin >> s >> t;
	s--, t--;
	queue<pair<int, int>> q;
	q.push({3 * s, 0});
	vector<int> visited(3 * n);
	while (!q.empty())
	{
		auto now = q.front();
		for (auto nxt : g[now.first])
		{
			if (!visited[nxt])
			{
				q.push({nxt, now.second + 1});
				visited[nxt] = now.second + 1;
			}
		}
		q.pop();
	}
	if (visited[3 * t])
	{
		cout << visited[3 * t] / 3 << endl;
	}
	else
	{
		cout << -1 << endl;
	}
}
```