#include "bits/stdc++.h"
using namespace std;

int N, M, S, T, INF = 1145141919;
static int Dis[100000][4] = {};
static vector<int> V[100000];

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; i++) for (int j = 0; j < 4; j++) Dis[i][j] = INF;
	for (int i = 0; i < M; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		V[u].push_back(v);
	}
	cin >> S >> T;
	S--, T--;
	Dis[S][0] = 0;
	queue<pair<int, int> > Q;
	for (int i : V[S]) Q.push({ i, 1 });
	while (!Q.empty()) {
		pair<int, int> p = Q.front();
		Q.pop();
		if (Dis[p.first][p.second % 3] != INF) continue;
		Dis[p.first][p.second % 3] = p.second;
		for (int next : V[p.first]) {
			if (Dis[next][(p.second + 1) % 3] == INF) {
				Q.push({ next, p.second + 1 });
			}
		}
	}
	if (Dis[T][0] == INF) {
		cout << -1;
		return 0;
	}
	cout << Dis[T][0] / 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 "bits/stdc++.h"
using namespace std;

int N, M, S, T, INF = 1145141919;
static int Dis[100000][4] = {};
static vector<int> V[100000];

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; i++) for (int j = 0; j < 4; j++) Dis[i][j] = INF;
	for (int i = 0; i < M; i++) {
		int u, v;
		cin >> u >> v;
		u--, v--;
		V[u].push_back(v);
	}
	cin >> S >> T;
	S--, T--;
	Dis[S][0] = 0;
	queue<pair<int, int> > Q;
	for (int i : V[S]) Q.push({ i, 1 });
	while (!Q.empty()) {
		pair<int, int> p = Q.front();
		Q.pop();
		if (Dis[p.first][p.second % 3] != INF) continue;
		Dis[p.first][p.second % 3] = p.second;
		for (int next : V[p.first]) {
			if (Dis[next][(p.second + 1) % 3] == INF) {
				Q.push({ next, p.second + 1 });
			}
		}
	}
	if (Dis[T][0] == INF) {
		cout << -1;
		return 0;
	}
	cout << Dis[T][0] / 3;
}
```