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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const int N=1e5+5,M=N<<2;
int ans,s,t,tot,now,n,m,head[N],nex[M],to[M];
int d[N];
vector<int>v[N];
void add(int u,int v)
{
    to[++tot]=v;nex[tot]=head[u];head[u]=tot;
}
queue<int>q;
void dfs(int u,int p)
{
    if(p==0)
    {
        if(u!=now)
            v[now].push_back(u);
        return;
    }
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];
        dfs(v,p-1);
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    for(int i=1;i<=n;i++)
        now=i,dfs(i,3);
    memset(d,inf,sizeof(d));
    scanf("%d%d",&s,&t);
    d[s]=0;
    q.push(s);
    while(!q.empty())
    {
        now=q.front();q.pop();
        if(d[now]+1>=d[t]) continue;
        for(int i=0;i<v[now].size();i++)
        {
            int y=v[now][i];
            if(d[y]>d[now]+1)
            {
                d[y]=d[now]+1;
                q.push(y);
            }
        }
    }
    if(d[t]!=d[0]) printf("%d\n",d[t]);
    else printf("-1\n");
}