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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,N) for(int i=0;i<int(N);++i)
#define rep1(i,N) for(int i=1;i<int(N);++i)
#define all(a) (a).begin(),(a).end()		//sort(all(vi S)) sort(all(string S))
#define print(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<_<<", "; cerr<<"]"<<endl; }
#define printpair(v) { cerr<<#v<<": [ "; for(auto _ : v) cerr<<"{"<<_.first<<","<<_.second<<"}"<<", "; cerr<<"]"<<endl; }

using P = pair<int, int>;		//P.first, P.second
typedef long long ll;
typedef vector<int> vi;
typedef set<int> seti;
typedef vector<string> vs;

const int MOD = 1e9+7;
const int INF = 1e9;

int dist[100005][3];
int main() {
	int N,M;
	cin>>N>>M;
	vector<vector<int>> to(N);
	rep(i,M){
		int a,b;
		cin>>a>>b;
		a--;b--;
		to[a].push_back(b);
	}
	int sv,tv;
	cin>>sv>>tv;
	sv--;tv--;
	rep(i,N)rep(j,3) dist[i][j] = INF;
	queue<P> q;
	q.push({sv,0});
	dist[sv][0] = 0;
	while(!q.empty()){
		int v = q.front().first;
		int l = q.front().second;
		q.pop();
		for(int u: to[v]){
			int nl = (l+1)%3;
			if(dist[u][nl]!=INF)continue;
			dist[u][nl] = dist[v][l] + 1;
			q.push({u,nl});
		}
	}
	int ans = dist[tv][0];
	if(ans == INF)ans = -1;
	else ans /= 3;
	cout<<ans<<endl;
}