#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int d[200000][3];
vector<int>E[200000];

int main(){
	int n,m;cin>>n>>m;
	rep(i,m){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
	}
	int s,t;cin>>s>>t;s--;t--;
	memset(d,-1,sizeof(d));
	queue<P>que;
	d[s][0]=0;
	que.push(P(s,0));
	while(!que.empty()){
		P p=que.front();que.pop();
		for(int u:E[p.first]){
			if(d[u][(p.second+1)%3]==-1){
				d[u][(p.second+1)%3]=d[p.first][p.second]+1;
				que.push(P(u,(p.second+1)%3));
			}
		}
	}
	if(d[t][0]==-1)puts("-1");
	else cout<<d[t][0]/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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

int d[200000][3];
vector<int>E[200000];

int main(){
	int n,m;cin>>n>>m;
	rep(i,m){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
	}
	int s,t;cin>>s>>t;s--;t--;
	memset(d,-1,sizeof(d));
	queue<P>que;
	d[s][0]=0;
	que.push(P(s,0));
	while(!que.empty()){
		P p=que.front();que.pop();
		for(int u:E[p.first]){
			if(d[u][(p.second+1)%3]==-1){
				d[u][(p.second+1)%3]=d[p.first][p.second]+1;
				que.push(P(u,(p.second+1)%3));
			}
		}
	}
	if(d[t][0]==-1)puts("-1");
	else cout<<d[t][0]/3<<endl;
}
```