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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define PI 3.14159265358979323846
#define int long long
constexpr long long INF = numeric_limits<long long>::max() / 2;
constexpr int MOD = 1000000007;
using Graph = vector<vector<int>>;



signed main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int N,M;
    cin>>N>>M;
    Graph G(N);

    int u[M],v[M];
    rep(i,M){
        cin>>u[i]>>v[i];
        u[i]--;
        v[i]--;
        G[u[i]].push_back(v[i]);
    }
    int S,T;
    cin>>S>>T;
    S--;
    T--;

    int dp[N][3];
    queue<pair<int,int>> q;
    q.push(make_pair(S,0));
    rep(i,N){
        rep(j,3){
            dp[i][j]=INF;
        }
    }
    dp[S][0]=0;
    while(!q.empty()){
        auto e=q.front();
        q.pop();
        int v=e.first;
        int d=e.second;
        for(auto nv:G[v]){
            int nd=(d+1)%3;
            if(dp[nv][nd]>dp[v][d]+1){
                dp[nv][nd]=dp[v][d]+1;
                q.push(make_pair(nv,nd));
            }
        }

    }
    if(dp[T][0]==INF){
        cout << -1 << endl;
    }else{
        cout << dp[T][0]/3 << endl;
    }


    



}