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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)


// Problem : E - Hopscotch Addict
// Contest : AtCoder - AtCoder Beginner Contest 132
// URL : https://atcoder.jp/contests/abc132/tasks/abc132_e
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

//#pragma GCC optimize "trapv"//  WA to RE int!!
#include <bits/stdc++.h>
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORR(i,n) for(int i=n;i>=0;i--)
#define pb push_back
#define vlli vector<int>
#define slli set<int>
#define mlli map<int,int>
#define int long long int 
#define test int tt1234; cin>>tt1234;while(tt1234--)
#define endl "\n"
#define all(zz) zz.begin(),zz.end()
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define error(x) cerr << #x <<" is " << x << endl;
#define Error(a,b) cerr<<"( "<<#a<<" , "<<#b<<" ) = ( "<<(a)<<" , "<<(b)<<" )\n";
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define fint fast;test
#define SI(a,n) int a[n];FOR(i,n) cin>>a[i];
#define input(a,n) int n;cin>>n; SI(a,n)
#define set(a,x) memset(a,x,sizeof(a))
#define L(x)	((x)<<1)
#define R(x)	(((x)<<1)+1)
#define ft first
#define se second
#define MOD 1000000007
using namespace std;
map<pair<int,int>,bool>edg;
vector<vector<int>>adj;
vector<bool>vis;
vector<int>temp;
void go(int node,int dis)
{
	if(dis==3)
	{
		if(vis[node])return ;
		vis[node]=true;
		temp.pb(node);
		return ;
	}
	for(auto child:adj[node])
	{
		go(child,dis+1);
	}
}
void solve(){
	int n,m;
	cin>>n>>m;
	adj.resize(n+1,vector<int>());
	vis.resize(n+1,false);
	FOR(i,m)
	{
		int u,v;
		cin>>u>>v;
		edg[{u,v}]=false;
		adj[u].pb(v);
	}
	int S,T;
	cin>>S>>T;
	queue<pair<int,int>>q;
	q.push({S,0});
	vis[S]=true;
	while(q.size())
	{
		auto Top=q.front();
		q.pop();
		int node=Top.ft,dis=Top.se;
		if(node==T)
		{
			cout<<(dis);
			return ;
		}
		temp.clear();
		go(node,0);
		for(auto child:temp)
		{
			// cerr<<child<<endl;
			q.push({child,dis+1});
		}
	}
	cout<<-1;
	return;	
}


int32_t main() {
    // fint{
    	fast;
    	solve();
    // }
    return 0;
}