#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
vector<int> a[N];
int b[N][3];
int main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < m; i++)
	{
		int u, v;
		cin >> u >> v;
		a[u].push_back(v);
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j < 3; j++)
		{
			b[i][j] = -3;
		}
	}
	int s, t;
	cin >> s >> t;
	queue <pair< int, int >> q;
	b[s][0] = 0;
	q.push({ s,0 });
	while (!q.empty())
	{
		const int cx = q.front().first, cy = q.front().second, ny = (cy + 1) % 3;
		q.pop();
		for (int i = 0; i<int(a[cx].size()); i++)
		{
			const int nx = a[cx][i];
			if (b[nx][ny] == -3)
			{
				q.push({ nx,ny });
				b[nx][ny] = b[cx][cy] + 1;
			}
		}
	}
	cout << b[t][0]/3 << endl;


}

### Prompt

Please formulate a cpp solution to the following problem:

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;
const int N = 1e5 + 10;
vector<int> a[N];
int b[N][3];
int main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < m; i++)
	{
		int u, v;
		cin >> u >> v;
		a[u].push_back(v);
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j < 3; j++)
		{
			b[i][j] = -3;
		}
	}
	int s, t;
	cin >> s >> t;
	queue <pair< int, int >> q;
	b[s][0] = 0;
	q.push({ s,0 });
	while (!q.empty())
	{
		const int cx = q.front().first, cy = q.front().second, ny = (cy + 1) % 3;
		q.pop();
		for (int i = 0; i<int(a[cx].size()); i++)
		{
			const int nx = a[cx][i];
			if (b[nx][ny] == -3)
			{
				q.push({ nx,ny });
				b[nx][ny] = b[cx][cy] + 1;
			}
		}
	}
	cout << b[t][0]/3 << endl;


}

```