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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,l,r) for(long long i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define REPS(i,n) FOR(i,1,n+1)
#define RFOR(i,l,r) for(long long i=(l);i>=(r);--i)
#define RREP(i,n) RFOR(i,N-1,0)
#define RREPS(i,n) RFOR(i,N,1)
#define int long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
const int INF=1e18;
const int MOD=1e9+7;
const int MAX_N=1e5;
typedef pair<int,int> P;
int N,M;
vector<int>Graph[MAX_N];
vector<int>depth;
queue<P>q;

void dfs(int k,int d,int p){
    if(p==3){
        if(depth[k]>d){
            depth[k]=d;
            q.push({k,d});
        }
        return;
    }
    for(auto v:Graph[k]){
        dfs(v,(p%3==0?d+1:d),p+1);
    }
}

signed main(){
    cin>>N>>M;
    REP(i,M){
        int a,b;cin>>a>>b;a--;b--;
        Graph[a].pb(b);
    }
  	REP(i,N)depth.pb(INF);
    int S,T;cin>>S>>T;S--;T--;
    depth[S]=0;
    q.push({S,0});
    while(q.size()){
        P p=q.front();q.pop();
        dfs(p.first,p.second,0);
    }
    cout<<(depth[T]==INF?-1:depth[T])<<endl;
}