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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.*;

class Main {
  public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      long cnt =0;
      long N = sc.nextInt();
      long K = sc.nextInt();
      
      for(long i=1;i<K+1;i++) {
    	  if (i == 1) {
        	  cnt = ( calcNumOfCombination(N-K+1,i) ) % 1000000007;    		  
    	  } else {
        	  cnt = ( calcNumOfCombination(N-K+1,i) * calcNumOfCombination(K-1,i-1) ) % 1000000007;
    	  }
    	  System.out.println(cnt);
      }
      

  }
  
  
 //組み合わせの数nCrを計算
  static long calcNumOfCombination(long n, long r){
	  double wk = 1;
	  for(long i=0; i<r; i++) {
		  wk = (wk * (n-i) / (r-i)) % 1000000007;
	  }
	  if(wk==0) wk=1;
    return (long) Math.round(wk);
}
//nの階乗を計算
static long factorial(long n){
    long answer = 1;
while(n > 1){
    answer *= n;
    n--;
}
    return answer;
}  
//nPrの計算
static long Parmitation(long n, long r){
	  double wk = 1;
	  for(long i=0; i<r+1; i++) {
		  wk = (wk * (n-i)) % 1000000007;
	  }
//   	  System.out.println("n=" + n + " r=" + r + " wk=" + wk);
   	  if(wk==0) wk=1;
   	   return (long) Math.round(wk);
}

}