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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

struct uvtyp {int u; int v;};

int asbc(const void *x, const void *y) {
  return (*(struct uvtyp *)x).u - (*(struct uvtyp *)y).u;
}

struct uvtyp uv[100000];
int m,d[100001],idx[100001];

void set3(int i, int k){
   int j;

   j=idx[i]; 
   if (j<0) return;
   for (;uv[j].u==i;j++) set(uv[j].v,k+1);
}

void set2(int i, int k){
   int j;

   j=idx[i]; 
   if (j<0) return;
   for (;uv[j].u==i;j++) set3(uv[j].v,k);
}

void set(int i, int k){
   int j;

   if (d[i]<0) d[i]=k;
   else if (d[i]<=k) return;
   else d[i]=k;
   j=idx[i]; 
   if (j<0) return;
   for (;uv[j].u==i;j++) set2(uv[j].v,k);
}

int main(){
   int i,j,n,s,t;
    
   scanf("%d %d",&n,&m);
   for (i=0;i<m;i++) scanf("%d %d",&(uv[i].u),&(uv[i].v));
   scanf("%d %d",&s,&t);
   if (m==0) {
     puts("-1");
     return 0;
   }
   for (i=1;i<=n;i++) idx[i]=d[i]=-1;
   qsort(uv,m,sizeof(struct uvtyp),asbc);
   for (i=0;i<m;i++)
       if (idx[uv[i].u]<0) idx[uv[i].u]=i;
   set(s,0);
   printf("%d\n",d[t]);
   return 0;
}