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 <iostream>
#include <cstring>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <math.h>
#include <string>
#include <map>
#include <cstdlib>
#include <sstream>
#include <bitset>
using namespace std;

typedef long long ll;
int mod=3;
int INF=100000000;
typedef pair<int, int> P;

int main()
{
    int N,M;
    cin >> N >> M;
    vector<int> vt[N];
    for(int i=0; i<M; i++){
        int t1,t2;
        cin >> t1 >> t2;
        t1--;
        t2--;
        vt[t1].push_back(t2);
    }
    int S,T;
    cin >> S >> T;
    S--;
    T--;

    int dist[N][3];
    for(int i=0; i<N; i++){
        for(int j=0; j<3; j++){
            dist[i][j]=INF;
        }
    }
    queue<P> que;
    que.push(P(S, 0));//今いる頂点、距離を3で割った余り
    while(!que.size()){
        int v=que.front().first;
        int l=que.front().second;
        que.pop();
        for(int u : vt[v]){
            int nl=(l+1)%3;
            if(dist[u][nl]!=INF) continue;
            dist[u][nl]=dist[v][l]+1;
            que.push(P(u,nl));
        }
    }
    int ans=dist[N-1][0];
    if(ans==INF) ans=-1;
    else ans/=3;
    cout << ans << endl;
    

    return 0;
}