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: Runtime Error)

#include<stdio.h>
int q[100010],ta[100010],nt[100010];
int MIN(int a,int b){return a<b?a:b;}
int topo(int v,int e,int *a,int *b){
  int i,t,r,c[100010]={0};
  for(i=0;i<e;i++)c[b[i]]++;
  for(i=r=0;i<v;i++){
    ta[i]=-1;
    if(c[i]==0)q[r++]=i;
  }
  for(i=0;i<e;i++){
    nt[i]=ta[a[i]];
    ta[a[i]]=i;
  }
  for(t=0;t<r;t++){
    for(i=ta[q[t]];i+1;i=nt[i]){
      if(--c[b[i]]==0)q[r++]=b[i];
    }
  }
  return r;
}
void mklsd(int v,int e,int *a,int *b){
  int i;
  for(i=0;i<v;i++)ta[i]=-1;
  for(i=0;i<e;i++){
    nt[i]=ta[a[i]];
    ta[a[i]]=i;
  }
}
int main(){
  int n,m,a[100010],b[100010],s,t,i,j,k,g,r;
  int c[100010][3];
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++){
    for(j=0;j<3;j++)c[i][j]=1e9;
  }
  for(i=0;i<m;i++){
    scanf("%d %d",&a[i],&b[i]);
    a[i]--;
    b[i]--;
  }
  scanf("%d %d",&s,&g);//printf("a\n");
  mklsd(n,m,a,b);
  /*for(i=0;i<n;i++){printf("%d:",i);
    for(j=ta[i];j+1;j=nt[j])printf("%d",b[j]);
    printf("\n");
    }///*/
  c[s-1][0]=0;
  q[t=0]=(s-1)*10+0;
  for(r=1;r-t;t++){
    j=(q[t]%10+1)%3;
    for(i=ta[q[t]/10];i+1;i=nt[i]){//printf("%d %d\n",q[t]/10,b[i]);
      if(c[b[i]][j]<1e8)continue;
      c[b[i]][j]=c[q[t]/10][q[t]%10]+1;
      q[r++]=b[i]*10+j;
    }
  }
  printf("%d\n",c[g-1][0]<1e8?c[g-1][0]/3:-1);
  return 0;
}