Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define N 100005
#define INF 1061109567
#define rep(i,n) for((i)=0;(i)<(n);(i)++)

int n,m;
int s,t;
vector<int> e[N];
int d[N][3];
priority_queue<pii,vector<pii>,greater<pii> > q;

int main()
{
	
	int i;
	scanf("%d%d",&n,&m);
	rep(i,m)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		u--; v--;
		e[u].pb(v);
	}
	scanf("%d%d",&s,&t);
	s--; t--;
	
	memset(d,63,sizeof(d));
	q.push(mp(0,s));
	while(!q.empty())
	{
		int v=q.top().second;
		int dist=q.top().first; 
		q.pop();
		if(dist>d[v][dist%3]) continue;
		rep(i,e[v].size())
		{
			int to=e[v][i];
			int &Di=d[to][(dist+1)%3];
			if(Di<=dist+1) continue;
			Di=dist+1;
			q.push(mp(Di,to));
		}
	}
	printf("%d\n",(d[t][0]!=INF)?(d[t][0]+2)/3:-1);
		
	return 0;
}