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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
int main(){
    long N,M,u,v,S,T;
    cin>>N>>M;
    vector<long>g[3*N],d(3*N);
    for(long i=0;i<M;i++){
        cin>>u>>v;
        u--,v--;
        g[3*u+0].push_back(3*v+1);
        g[3*u+1].push_back(3*v+2);
        g[3*u+2].push_back(3*v+0);
    }
    cin>>S>>T;
    S--,T--;
    for(long i=0;i<3*N;i++)d[i]=100000000000000L;
    d[0]=0L;
    queue<pair<long,long>>q;
    q.push(make_pair(3*S,0L));
    while(!q.empty()){
        auto c=q.front();q.pop();
        if(c.first==3L*T){cout<<c.second/3L<<endl;return 0;}
        for(auto x:g[c.first])if(d[x]==100000000000000L)d[x]=c.second+1L,q.push(make_pair(x,c.second+1L));
    }
    cout<<-1<<endl;
    return 0;
}