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::literals::string_literals;
using i64 = long long;
using std::cout;
using std::endl;
using std::cin;

template<typename T>
std::vector<T> make_v(size_t a){return std::vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return std::vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

int main() {
	int n, m; scanf("%d%d", &n, &m);
	std::vector<std::vector<int>> g(n);
	for(int i = 0; i < m; i++) {
		int a, b; scanf("%d%d", &a, &b);
		
		g[a - 1].push_back(b - 1);
	}
	int s, t; scanf("%d%d", &s, &t); s--; t--;
	
	auto min_cost = make_v<int>(n, 4);
	for(auto &w: min_cost) for(auto &v: w) v = 1 << 30; min_cost[s][0] = 0;
	std::queue<std::pair<int, std::pair<int, int>>> qu; qu.push({s, {3, 1}});
	while(!qu.empty()) {
		auto p = qu.front(); qu.pop();
		
		for(auto e: g[p.first]) {
			int c = p.second.first - 1;
			if(min_cost[e][c] <= p.second.second) continue;
			min_cost[e][c] = p.second.second;
			if(c == 0) qu.push({e, {3, p.second.second + 1}});
			else qu.push({e, {c, p.second.second}});
		}
	}
	
	if(min_cost[t][0] == 1 << 30) printf("-1\n");
	else printf("%d\n", min_cost[t][0]);
	return 0;
}