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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;
struct Edge {
	ll from;
	ll to;
	ll cost;
};
using Graph = vector<vector<Edge>>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(a) (a).begin(), (a).end()
#define RES(a, n) (a).resize(n)
#define VLCIN(a, n) REP(i,n){cin >> a[i];}
/***** define constant start *******/
const ll MOD = 1000000007;
const double PI = 2.0 * asin(1.0);
const ll INF = 1L<<60;
string abc = "abcdefghijklmnopqrstuvwxyz";
string ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll MAX = (ll)pow(10,7);
/****** define constant end ********/
/****** define variable start ******/
ll n, m;
vl u, v;
ll s, t;
vll G;
vll Gans;
/******* define variable end *******/
void dfs(ll v, ll one, ll two, ll three){
	if(three != -1){
		auto ite = lower_bound(ALL(Gans[three]), v);
		if (ite != Gans[three].end())if(*ite == v){
			return;
		}
		Gans[three].push_back(v);
	}
	for(auto nv : G[v]){
		dfs(nv, v,one, two);
	}
	return;
}
int solve() {
	cin >> n >> m;
	RES(Gans, n);
	RES(u, m);
	RES(v, m);
	RES(G, n);
	REP(i, m){
		cin >> u[i] >> v[i];
		G[u[i]-1].push_back(v[i]-1);
	}
	cin >> s >> t;
	dfs(0, -1, -1, -1);
	  // BFS のためのデータ構造
    vector<ll> dist(n, -1); // 全頂点を「未訪問」に初期化
    queue<ll> que;

    // 初期条件 (頂点 0 を初期ノードとする)
    dist[s-1] = 0;
    que.push(s-1); // 0 を橙色頂点にする

    // BFS 開始 (キューが空になるまで探索を行う)
    while (!que.empty()) {
        ll v = que.front(); // キューから先頭頂点を取り出す
        que.pop();
		if(v == t-1){
			cout << dist[t-1] << endl;
			return 0;
		}

        // v から辿れる頂点をすべて調べる
        for (ll nv : Gans[v]) {
            if (dist[nv] != -1) continue; // すでに発見済みの頂点は探索しない

            // 新たな白色頂点 nv について距離情報を更新してキューに追加する
            dist[nv] = dist[v] + 1;
            que.push(nv);
        }
    }
	cout << dist[t-1] << endl;
	return 0;
}
int main() {
	//clock_t start = clock();
	cout << std::fixed << std::setprecision(10);
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int end_solve = solve();
	/*
	clock_t end = clock();
	const double time = static_cast<double>(end - start) / CLOCKS_PER_SEC *
	1000.0; printf("time %lf[ms]\n", time);
	*/

	return end_solve;
}