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;

typedef long long ll;
typedef unsigned long long ull;

template <typename T, typename C>
std::array<std::vector<T>,2> Dijkstra(std::vector<C> g, T istart){
    auto n = g.size();
    priority_queue<std::array<T,2>> q;
    std::vector<T> d(n, numeric_limits<T>::max());
    std::vector<T> prev(n, numeric_limits<T>::max());

    d[istart] = 0;
    q.push(std::array<T,2>{d[istart], istart});

    while(!q.empty()){
        auto u = q.top()[1]; q.pop();
        for(auto v : g[u]){
            auto alt = d[u] + 1;
            if(alt < d[v]){
                d[v] = alt;
                prev[v] = u;
                q.push(std::array<T,2>{-alt, v});
            }
        }
    }
    return std::array<std::vector<T>,2>{d, prev};
}

int main()
{
    ll N, M;
    cin >> N >> M;
    ll ui, vi;
    ll S, T;
    vector<set<ll>> G(3*N);
    for(int i=0;i<M;i++){
        cin >> ui >> vi;
        ui--; vi--;
        G[3*ui+0].insert(3*vi+1);
        G[3*ui+1].insert(3*vi+2);
        G[3*ui+2].insert(3*vi+0);
    }
    cin >> S >> T;
    S--; T--;

    auto r = Dijkstra(G, 3*S);

    if(r[0][3*T]==numeric_limits<ll>::max()){
        cout<<"-1"<<endl;
    }else{
        cout<<r[0][3*T]/3<<endl;
    }

    return 0;
}