#include <bits/stdc++.h>
using namespace std;
using lint = long long;
const lint INF = 1e18;
typedef pair<lint, lint> P;

signed main(){
	lint N, M; cin >> N >> M;
	vector<vector<lint> > G(N);
	for(lint i = 0; i < M; i++){
		lint u, v; cin >> u >> v; --u; --v;
		G[u].push_back(v); 
	}
	lint s, t; cin >> s >> t; --s; --t;
	vector<vector<lint> > dist(N, vector<lint> (3, INF));
	queue<P> que;
	que.push(P(s, 0)); dist[s][0] = 0;

	while(que.size()){
		P p = que.front(); que.pop();
		lint now = p.first, num = p.second;
		for(auto next : G[now]){
			lint num_next = (num + 1) % 3;
			if(dist[next][num_next] == INF){
				dist[next][num_next] = dist[now][num] + 1;
				que.push(P(next, num_next));
			}
		}
	}

	lint ans = dist[t][0];
	if(ans == INF) cout << -1 << endl;
	else cout << ans/3 << endl;
}
### Prompt

Your task is to create 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;
using lint = long long;
const lint INF = 1e18;
typedef pair<lint, lint> P;

signed main(){
	lint N, M; cin >> N >> M;
	vector<vector<lint> > G(N);
	for(lint i = 0; i < M; i++){
		lint u, v; cin >> u >> v; --u; --v;
		G[u].push_back(v); 
	}
	lint s, t; cin >> s >> t; --s; --t;
	vector<vector<lint> > dist(N, vector<lint> (3, INF));
	queue<P> que;
	que.push(P(s, 0)); dist[s][0] = 0;

	while(que.size()){
		P p = que.front(); que.pop();
		lint now = p.first, num = p.second;
		for(auto next : G[now]){
			lint num_next = (num + 1) % 3;
			if(dist[next][num_next] == INF){
				dist[next][num_next] = dist[now][num] + 1;
				que.push(P(next, num_next));
			}
		}
	}

	lint ans = dist[t][0];
	if(ans == INF) cout << -1 << endl;
	else cout << ans/3 << endl;
}
```