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: Runtime Error)

#include <bits/stdc++.h>
#define mem(dp,a) memset(dp,a,sizeof(dp))
#define pb(x) push_back(x)
#define m_p(x,y) make_pair(x,y)
#define rep(i,a,b) for(ll i=a;i<b;i++)
#define repush_back(i,a,b) for(ll i=a;i>=b;i--)
#define f(n) for(ll i=0;i<n;i++)
#define r(n) for(ll j=0;j<n;j++)
#define F first
#define S second
#define pi 3.14159265359
#define hs ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
typedef long long int ll;
ll HRX=1e18;
ll INF=1e9+7;

const ll N=2*100001;
vector<ll>adj[N];
bool vis[N];
ll n,e,src,dest;
ll dist[N];

void bfs(ll s)
{
  
  
  queue<ll>q;
  if(!vis[s])
    vis[s]=true;
  q.push(s);
  dist[s]=0;
  while(!q.empty())
  {
    ll u=q.front();
    q.pop();
    f(adj[u].size())
    {
     ll v=adj[u][i];
     if(!vis[v])
     {
        vis[v]=1;
        dist[v]=dist[u]+1;
        q.push(v);
     }
    }
  }
  if(dist[dest]%3==0 && dist[dest])
    cout<<(dist[dest]/3)<<endl;
  else
    cout<<"-1"<<endl;

}
int main()
{
  hs;
  cin>>n>>e;
  f(e)
  {
     ll x,y;
     cin>>x>>y;
     adj[x].pb(y+n);
     adj[x+n].pb(y+2*n);
     adj[x+2*n].pb(y);
  }
  cin>>src>>dest;
  mem(vis,0);
  bfs(src);
  return 0;


}