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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

const int infI=1e9+5;
const long long infL=1e18;
const int MOD=1e9+7;
const int MAX=1e5+5;

#define mp make_pair
#define in &
#define pb push_back
#define ii pair<int,int>
#define iii pair<ii,int>
#define ll long long
#define pll pair<ll,ll>
#define vl vector<vector<ll> >
#define vi vector<vector<int> >
#define eb emplace_back
#define forn(i,a,b) for(int i=a;i<b;i++)
#define fastio ios_base::sync_with_stdio(0);cin.tie(0)
#define endl '\n'
#define popb pop_back();
#define se second
#define fi first

vi adj;
int n,m,s,d;
/*
    we create 3 copies of this graph and for every vertex u, we create u1,u2,u3 and for every edge (u,v), we create
    (u1,v2),(u2,v3), and (u3,v1). Look at this as if we have 3 layers and we move from one layer to another.
    If we can move from s to t through the other layers and this is the shortest path , then this path is also divisible 
    by 3, as we move through 3 edges to get back to same layer.In this implementation,we
    have a directed edge from layer 1->2, 2->3 and 3->1.we can also connect layers 1->3,3->2 and 2->1.
    We just can't connect a layer with itself.
*/
int main(){
    scanf("%d%d",in n,in m);
    adj.resize(3*n+1);
    forn(i,0,m){
        int s,d;
        scanf("%d%d",in s,in d);
        adj[s].pb(d+n);
        adj[s+n].pb(d+2*n);
        adj[s+2*n].pb(d);
    }
    int s,d;
    scanf("%d%d",in s,in d);
    ll dist[3*n+1];
    bool vis[3*n+1]={};
    memset(dist,infI,sizeof(dist));
    dist[s]=0;
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=true;
        for(auto it:adj[u]){
            if(!vis[it]){
                dist[it]=dist[u]+1;
                vis[it]=true;
                q.push(it);
            }
        }
    }
    if(dist[d]!=infI){
        printf("%lld\n",dist[d]/3);
    }
    else{
        printf("-1\n");
    }
    return 0;
}