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: Accepted)

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

long *tree[300001];

int main(void) {

  long n,m;
  scanf("%ld %ld", &n, &m);
  long tree_s[n*3+1];
  for (long i = 0; i <= n*3; i++) {
    tree_s[i] = 0;
  }
  long u,v;
  for (long i = 0; i < m; i++) {
    scanf("%ld %ld", &u, &v);
    tree[u] = realloc(tree[u], sizeof(long)*(tree_s[u]+1));
    tree[u][tree_s[u]] = n+v;
    tree_s[u]++;
    tree[n+u] = realloc(tree[n+u], sizeof(long)*(tree_s[n+u]+1));
    tree[n+u][tree_s[n+u]] = n*2+v;
    tree_s[n+u]++;
    tree[n*2+u] = realloc(tree[n*2+u], sizeof(long)*(tree_s[n*2+u]+1));
    tree[n*2+u][tree_s[n*2+u]] = v;
    tree_s[n*2+u]++;
  }
  long s,t;
  scanf("%ld %ld", &s, &t);
  long visited[n*3+1];
  for (long i = 0; i <= n*3; i++) {
    visited[i] = -1;
  }
  visited[s] = 0;
  long queue[n*3];
  queue[0] = s;
  long start = 0, end = 0, tail = 0;
  long cur,next;
  int flag;
  while (1) {
    flag = 0;
    for (long i = start; i <= end; i++) {
      cur = queue[i];
      for (long j = 0; j < tree_s[cur]; j++) {
        next = tree[cur][j];
        if (visited[next] != -1) {
          continue;
        }
        visited[next] = visited[cur]+1;
        tail++;
        queue[tail] = next;
        flag = 1;
      }
    }
    if (flag == 0 || tail == n*3-1) {
      break;
    }
    start = end+1;
    end = tail;
  }
  if (visited[t] == -1) {
    printf("-1\n");
  } else {
    printf("%ld\n", visited[t]/3);
  }

  return 0;
}