#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int INF=1<<29;

int main(){
	int n,m; scanf("%d%d",&n,&m);
	vector<int> G[100000];
	rep(i,m){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		G[u].emplace_back(v);
	}
	int s,t; scanf("%d%d",&s,&t); s--; t--;

	int d[100000][3];
	rep(u,n) rep(k,3) d[u][k]=INF;
	d[s][0]=0;
	priority_queue<tuple<int,int,int>> Q;
	Q.emplace(0,s,0);
	while(!Q.empty()){
		auto a=Q.top(); Q.pop();
		int d0=-get<0>(a),u=get<1>(a),k=get<2>(a);
		if(d0>d[u][k]) continue;

		if(u==t && k==0){ printf("%d\n",d0/3); return 0; }

		for(int v:G[u]){
			if(d[v][(k+1)%3]>d0+1){
				d[v][(k+1)%3]=d0+1;
				Q.emplace(-d[v][(k+1)%3],v,(k+1)%3);
			}
		}
	}

	puts("-1");

	return 0;
}

### Prompt

In CPP, your task is to solve 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++)

using namespace std;

const int INF=1<<29;

int main(){
	int n,m; scanf("%d%d",&n,&m);
	vector<int> G[100000];
	rep(i,m){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		G[u].emplace_back(v);
	}
	int s,t; scanf("%d%d",&s,&t); s--; t--;

	int d[100000][3];
	rep(u,n) rep(k,3) d[u][k]=INF;
	d[s][0]=0;
	priority_queue<tuple<int,int,int>> Q;
	Q.emplace(0,s,0);
	while(!Q.empty()){
		auto a=Q.top(); Q.pop();
		int d0=-get<0>(a),u=get<1>(a),k=get<2>(a);
		if(d0>d[u][k]) continue;

		if(u==t && k==0){ printf("%d\n",d0/3); return 0; }

		for(int v:G[u]){
			if(d[v][(k+1)%3]>d0+1){
				d[v][(k+1)%3]=d0+1;
				Q.emplace(-d[v][(k+1)%3],v,(k+1)%3);
			}
		}
	}

	puts("-1");

	return 0;
}

```