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: Runtime Error)

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
using Graph = vector<vector<int> >;

int main(){
    int n, m;
    cin >> n >> m;
    Graph G(n+1);
    int v;
    int u;
    for(int i = 1; i <= m; i++){
      cin >> u >> v;
      G[u].push_back(v);
    }
    int s, t;
    cin >> s >> t;
    bool isVisit[n+1][n+1];
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            isVisit[i][j] = false;
        }
    }

    queue<pair<int, int> > queue;
    for(int i = 0; i < G[s].size(); i++){
        pair<int, int> p = make_pair(G[s][i], 1);
        queue.push(p);
    }
    int ans = -1;
    int count = 0;
    while(!queue.empty()){
        bool isEnd = false;
        pair<int, int> next = queue.front();
        int node = next.first;
        int depth = next.second;
        if(depth % 3 == 0){
            isVisit[node][depth / 3] = true;
        }
        queue.pop();
        if(node == t && depth % 3 == 0){
            ans = depth / 3;
            break;
        }
        for(int i = 0; i < G[node].size(); i++){
            int node_next = G[node][i];
            int depth_next = depth + 1;
            pair<int, int> p = make_pair(node_next, depth_next);
            if(depth_next % 3 == 0){
                if(isVisit[node_next][depth_next / 3]){
                } else {
                    queue.push(p);
                }
            } else {
                queue.push(p);
            }
        }
        if(depth == 3*n + 1){
            ans = -1;
            break;
        }
    }
    cout << ans << endl;
    return 0;
}