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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
typedef long long ll;

ll n,m,u,v,s,t,d[100009][3];
vector<ll> e[100009];

int main(){
    cin>>n>>m;
    rep(i,n)rep(j,3)d[i+1][j]=1e18;
    rep(i,m){
        cin>>u>>v;
        e[u].push_back(v);
    }
    cin>>s>>t;
    queue<pair<int,int>> que;
    que.push({s,0});
    d[s][0]=0;
    while(!que.empty()){
        int v=que.front().first,x=que.front().second;
        que.pop();
        int y=(x+1)%3;
        rep(i,e[v].size()){
            if(d[e[v][i]][y]==1e18){
                que.push({e[v][i],y});
                d[e[v][i]][y]=d[v][x]+1;
            }
        }
    }
    cout<<(d[t][0]!=1e18?d[t][0]/3:-1)<<endl;
}