#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for (int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
#define f first
#define s second

int n, m, a, b, dist[100050][3], st, t;
vector<int> adj[100050];
queue<pi> q;

int main() {
	cin >> n >> m;
	FOR(i, 1, m) {
		cin >> a >> b;
		adj[a].push_back(b);
	}
	cin >> st >> t;
	q.push(pi(st, 0));
	memset(dist, -1, sizeof dist);
	dist[st][0] = 0;
	while (!q.empty()) {
		pi x = q.front(); q.pop();
		int nd = (x.s + 1) % 3;
		for (auto it:adj[x.f]) {
			if (dist[it][nd] != -1) continue;
			dist[it][nd] = dist[x.f][x.s] + (x.s == 0);
			q.push(pi(it, nd));
		}
	}
	cout << dist[t][0];
}


### 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;
#define FOR(i, a, b) for (int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for (int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
#define f first
#define s second

int n, m, a, b, dist[100050][3], st, t;
vector<int> adj[100050];
queue<pi> q;

int main() {
	cin >> n >> m;
	FOR(i, 1, m) {
		cin >> a >> b;
		adj[a].push_back(b);
	}
	cin >> st >> t;
	q.push(pi(st, 0));
	memset(dist, -1, sizeof dist);
	dist[st][0] = 0;
	while (!q.empty()) {
		pi x = q.front(); q.pop();
		int nd = (x.s + 1) % 3;
		for (auto it:adj[x.f]) {
			if (dist[it][nd] != -1) continue;
			dist[it][nd] = dist[x.f][x.s] + (x.s == 0);
			q.push(pi(it, nd));
		}
	}
	cout << dist[t][0];
}


```