#include <bits/stdc++.h>
using namespace std;
#define REP(i,a) for(int i = 0; i < (a); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;

int main(){
	int n,m;
	cin >> n >> m;
	int u,v;
	vector<vector<int>> to(n);
	REP(i,m){
		cin >> u >> v;
		u--;
		v--;
		to[u].push_back(v);
	}
	int s,t;
	cin >> s >> t;
	s--;
	t--;
	int dist[n][3];
	REP(i,n) REP(j,3) dist[i][j] = INF;
	queue<P> que;
	que.push(P(s,0));
	dist[s][0] = 0;
	while(que.size()){
		P p = que.front();
		que.pop();
		for(int i : to[p.first]){
			if(dist[i][(p.second + 1) % 3] == INF){
				dist[i][(p.second + 1) % 3] = dist[p.first][p.second] + 1;
				que.push(P(i, (p.second + 1) % 3));
			}
		}
	}
	if(dist[t][0] == INF) cout << -1 << endl;
	else cout << dist[t][0] / 3 << endl;
}
### 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;
#define REP(i,a) for(int i = 0; i < (a); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;

int main(){
	int n,m;
	cin >> n >> m;
	int u,v;
	vector<vector<int>> to(n);
	REP(i,m){
		cin >> u >> v;
		u--;
		v--;
		to[u].push_back(v);
	}
	int s,t;
	cin >> s >> t;
	s--;
	t--;
	int dist[n][3];
	REP(i,n) REP(j,3) dist[i][j] = INF;
	queue<P> que;
	que.push(P(s,0));
	dist[s][0] = 0;
	while(que.size()){
		P p = que.front();
		que.pop();
		for(int i : to[p.first]){
			if(dist[i][(p.second + 1) % 3] == INF){
				dist[i][(p.second + 1) % 3] = dist[p.first][p.second] + 1;
				que.push(P(i, (p.second + 1) % 3));
			}
		}
	}
	if(dist[t][0] == INF) cout << -1 << endl;
	else cout << dist[t][0] / 3 << endl;
}
```