#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 3e5 + 10;
int dis[MAXN], n, m, S, T;
vector<int> Mat[MAXN];
bool se[MAXN];
queue<int> BFS;

void addv(int v) {
	int a = v / 3;
	int nxt = v % 3 + 1;
	nxt %= 3;
	for (int u : Mat[a]) {
		int t = 3 * u + nxt;
		if (se[t])
			continue;
		se[t] = 1;
		BFS.push(t);
		dis[t] = dis[v] + 1;
	}
}

int main() {
	for (int i = 0; i < MAXN; i++)
		dis[i] = -3;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		Mat[a].push_back(b);
	}
	cin >> S >> T;
	dis[3 * S] = 0;
	se[3 * S] = 1;
	BFS.push(3 * S);
	while (!BFS.empty())
		addv(BFS.front()), BFS.pop();
	cout << dis[3 * T] / 3;
}
### Prompt

Please create a solution in CPP 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 <iostream>
#include <vector>
#include <queue>
using namespace std;

const int MAXN = 3e5 + 10;
int dis[MAXN], n, m, S, T;
vector<int> Mat[MAXN];
bool se[MAXN];
queue<int> BFS;

void addv(int v) {
	int a = v / 3;
	int nxt = v % 3 + 1;
	nxt %= 3;
	for (int u : Mat[a]) {
		int t = 3 * u + nxt;
		if (se[t])
			continue;
		se[t] = 1;
		BFS.push(t);
		dis[t] = dis[v] + 1;
	}
}

int main() {
	for (int i = 0; i < MAXN; i++)
		dis[i] = -3;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		Mat[a].push_back(b);
	}
	cin >> S >> T;
	dis[3 * S] = 0;
	se[3 * S] = 1;
	BFS.push(3 * S);
	while (!BFS.empty())
		addv(BFS.front()), BFS.pop();
	cout << dis[3 * T] / 3;
}
```