
#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=1e18;

vector<ll> g[100005];
ll u[3][100005];



int main() {
  ll n,m;
  cin >> n>>m;
  rep(i,m){
    ll u1,v1;
    cin>>u1>>v1;
    u1--;v1--;
    g[u1].push_back(v1);
  //  g[v1].push_back(u1);

  }
  rep(i,3)rep(j,100005)u[i][j]=-3;
  ll s1,t1;
  cin>>s1>>t1;
  s1--;t1--;
  queue <P> q;
  q.push(make_pair(s1,0));
  while(!q.empty()){
    P now=q.front();
    q.pop();
    ll t=now.first;
    ll dist=now.second;
    for(ll to:g[t]){
      if (u[(dist+1)%3][to]!=-3)continue;
  //    cout<<"to"<<to<<"d"<<(dist+1)<<endl;
      u[(dist+1)%3][to]=dist+1;
      q.push(make_pair(to,dist+1));
    }
  }

cout<<u[0][t1]/3;

}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp

#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=1e18;

vector<ll> g[100005];
ll u[3][100005];



int main() {
  ll n,m;
  cin >> n>>m;
  rep(i,m){
    ll u1,v1;
    cin>>u1>>v1;
    u1--;v1--;
    g[u1].push_back(v1);
  //  g[v1].push_back(u1);

  }
  rep(i,3)rep(j,100005)u[i][j]=-3;
  ll s1,t1;
  cin>>s1>>t1;
  s1--;t1--;
  queue <P> q;
  q.push(make_pair(s1,0));
  while(!q.empty()){
    P now=q.front();
    q.pop();
    ll t=now.first;
    ll dist=now.second;
    for(ll to:g[t]){
      if (u[(dist+1)%3][to]!=-3)continue;
  //    cout<<"to"<<to<<"d"<<(dist+1)<<endl;
      u[(dist+1)%3][to]=dist+1;
      q.push(make_pair(to,dist+1));
    }
  }

cout<<u[0][t1]/3;

}

```