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)

// 7/3
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); ++i)
using ll = long long;
const int INF = 1000000007;

// 頂点の状態を管理する。pair<最短距離, pair<頂点番号、到達時間のMOD3>>
using P = pair<int, pair<int, int>>;
using Graph = vector<vector<int>>;

int mod = 3;

int main() {
    int N, M;
    cin >> N >> M;

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

    int S, T;  // start, end
    cin >> S >> T;
    S--; T--;

    // dist[v][k] ... 頂点vにp(mod)回で到達する際の最短距離
    vector<vector<int>> dist(N, vector<int>(mod, INT_MAX));

    // 初期値のセット
    dist[S][0] = 0;

    // 最短頂点候補を保持するキュー
    priority_queue<P, vector<P>, greater<P>> que;
    que.push(P(0, {S, 0}));

    // 拡張ディクストラ
    while (!que.empty()) {
        auto p = que.top();
        que.pop();

        int v = p.second.first;
        int k = p.second.second;

        if (dist[v][k] < p.first) continue;

        for (auto nv : G[v]) {
            int nk = (k + 1) % mod;
            if (dist[nv][nk] > dist[v][k] + 1) {
                dist[nv][nk] = dist[v][k] + 1;
                que.push(P(dist[nv][nk], {nv, nk}));
            }
        }
    }

    if (dist[T][0] != INT_MAX) cout << dist[T][0] / 3 << endl;
    else cout << -1 << endl;
}