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

int N, M, S, T, A[1 << 18], B[1 << 18], dp[1 << 18][3];
vector<int> X[1 << 18];

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> A[i] >> B[i];
		X[A[i]].push_back(B[i]);
	}
	cin >> S >> T;
	for (int i = 1; i <= N; i++) { dp[i][0] = (1 << 30); dp[i][1] = (1 << 30); dp[i][2] = (1 << 30); }

	queue<pair<int, int>> Q;
	dp[S][0] = 0; Q.push(make_pair(S, 0));

	while (!Q.empty()) {
		int pos1 = Q.front().first, pos2 = Q.front().second; Q.pop();

		for (int to : X[pos1]) {
			if (dp[to][(pos2 + 1) % 3] > dp[pos1][pos2] + 1) {
				dp[to][(pos2 + 1) % 3] = dp[pos1][pos2] + 1;
				Q.push(make_pair(to, (pos2 + 1) % 3));
			}
		}
	}

	if (dp[T][0] == (1 << 30)) cout << "-1" << endl;
	else cout << dp[T][0] / 3 << endl;
	return 0;
}
### Prompt

Develop a solution in CPP 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 <iostream>
#include <queue>
using namespace std;

int N, M, S, T, A[1 << 18], B[1 << 18], dp[1 << 18][3];
vector<int> X[1 << 18];

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> A[i] >> B[i];
		X[A[i]].push_back(B[i]);
	}
	cin >> S >> T;
	for (int i = 1; i <= N; i++) { dp[i][0] = (1 << 30); dp[i][1] = (1 << 30); dp[i][2] = (1 << 30); }

	queue<pair<int, int>> Q;
	dp[S][0] = 0; Q.push(make_pair(S, 0));

	while (!Q.empty()) {
		int pos1 = Q.front().first, pos2 = Q.front().second; Q.pop();

		for (int to : X[pos1]) {
			if (dp[to][(pos2 + 1) % 3] > dp[pos1][pos2] + 1) {
				dp[to][(pos2 + 1) % 3] = dp[pos1][pos2] + 1;
				Q.push(make_pair(to, (pos2 + 1) % 3));
			}
		}
	}

	if (dp[T][0] == (1 << 30)) cout << "-1" << endl;
	else cout << dp[T][0] / 3 << endl;
	return 0;
}
```