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 <iostream>
#include <queue>
#include <vector>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;

    vector<vector<int>> graph(3 * n, vector<int>(0));
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;

        graph[u - 1].push_back(n + v - 1);
        graph[n + u - 1].push_back(2 * n + v - 1);
        graph[2 * n + u - 1].push_back(v - 1);
    }

    int s, t;
    cin >> s >> t;

    long long inf = 1LL << 48;
    vector<long long> dist(3 * n, inf);
    dist[s - 1] = 0;

    priority_queue<pair<long long, int>> que;
    que.push(make_pair(0, s - 1));
    while (!que.empty()) {
        pair<long long, int> cur = que.top();
        que.pop();

        for (int i = 0; i < static_cast<int>(graph[cur.second].size()); i++) {
            if (dist[graph[cur.second][i]] > dist[cur.second] + 1) {
                dist[graph[cur.second][i]] = dist[cur.second] + 1;
                que.push(make_pair(dist[cur.second] + 1, graph[cur.second][i]));
            }
        }
    }

    cout << (dist[t - 1] != inf ? dist[t - 1] / 3 : -1) << endl;
}