#include <bits/stdc++.h>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7;

vi G[300005];
int vis[300005];

int main()
{
	IO_OP;
	
	int n, m;
	cin >> n >> m;
	for(int i=0;i<m;i++) {
		int u, v;
		cin >> u >> v;
		G[u*3].PB(v*3+1);
		G[u*3+1].PB(v*3+2);
		G[u*3+2].PB(v*3);
	}
	int s, t;
	cin >> s >> t;
	s = s * 3;
	t = t * 3;
	vector<pi> v({{s, 0}});
	vis[s] = 1;
	for(int i=0;i<v.size();i++) {
		if(v[i].F == t) {
			cout << v[i].S/3 << endl;
			return 0;
		}
		for(int u:G[v[i].F]) {
			if(!vis[u]) {
				vis[u] = 1;
				v.EB(u, v[i].S+1);
			}
		}
	}
	cout << -1 << 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>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define EB emplace_back

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;

const int INF = 1e9 + 7;

vi G[300005];
int vis[300005];

int main()
{
	IO_OP;
	
	int n, m;
	cin >> n >> m;
	for(int i=0;i<m;i++) {
		int u, v;
		cin >> u >> v;
		G[u*3].PB(v*3+1);
		G[u*3+1].PB(v*3+2);
		G[u*3+2].PB(v*3);
	}
	int s, t;
	cin >> s >> t;
	s = s * 3;
	t = t * 3;
	vector<pi> v({{s, 0}});
	vis[s] = 1;
	for(int i=0;i<v.size();i++) {
		if(v[i].F == t) {
			cout << v[i].S/3 << endl;
			return 0;
		}
		for(int u:G[v[i].F]) {
			if(!vis[u]) {
				vis[u] = 1;
				v.EB(u, v[i].S+1);
			}
		}
	}
	cout << -1 << endl;
}


```