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 <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <assert.h>
typedef int64_t ll;
typedef uint64_t ull;
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (100000)
#define MOD (1000000007)


//グラフ
#define MAXV (MAXN*3)
int Gv;
typedef struct edge
{
    int to;
    ll cost;
    struct edge* next;
}edge;


void Gadd(edge** es,int v, int to, ll cost)
{
    edge* e = malloc(sizeof(edge));
    e->to = to;
    e->cost = cost;
    e->next = es[v];
    es[v] = e;
}

// Que
typedef struct Que
{
    ll val;
    struct Que* next;
}Que;

void QuePush(Que** qr,ll val)
{
    Que* q = malloc(sizeof(Que));
    q->val = val;
    q->next = *qr;
    *qr = q;
}

//NULLチェックは外側で
ll QuePop(Que** qr)
{
    ll r = (*qr)->val;
    Que* f = *qr;
    *qr = (*qr)->next;
    free(f);
    return r;
}

edge* Ge[MAXV]; //vから出てる有向辺
int dr[MAXV];
int ans = INT_MAX;
int s,t;

void bfs(void)
{
    Que *q=NULL,*nq=NULL;
    QuePush(&q,s);
    dr[s] = 0;
    while(q!=NULL)
    {
        while(q!=NULL)
        {
            ll v = QuePop(&q);
            edge* e = Ge[v];
            while(e!=NULL)
            {
                if(dr[e->to]==0)
                {
                    dr[e->to] = dr[v]+1;
                    QuePush(&nq,e->to);
                }
                e = e->next;
            }
        }
        q = nq;
        nq = NULL;
    }
}

int main(void)
{
    int n,m;
    scanf("%d %d",&n,&m);

    for(int i=0;i<m;i++)
    {
        int u,v;
        scanf("%d %d",&u,&v);
        u--;v--;
        Gadd(Ge,u,v+MAXN,0);
        Gadd(Ge,u+MAXN,v+MAXN+MAXN,0);
        Gadd(Ge,u+MAXN+MAXN,v,0);
    }
    scanf("%d %d",&s,&t);
    s--;t--;
    bfs();

    printf("%d\n",dr[t]==0?-1:dr[t]/3);

    return 0;
}