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

const int MOD=1000000007;
const int INF=1001001001;


int main(){
  int n,m;
  cin>>n>>m;
  vector<vector<int>> to(n);
  rep(i,m){
    int u,v;

    cin>>u>>v;
    u--;v--;
    to[u].push_back(v);
  }
  int s,t;
  cin>>s>>t;
  s--;t--;
  int dist[n][3];
  rep(i,n)rep(j,3)dist[i][j]=INF;
  dist[s][0]=0;
  queue<P> q;
  q.push(P(s,0));
  while (!q.empty()){
    int t1,t2;
    t1=q.front().first;
    t2=q.front().second;
    q.pop();
    for(int t3:to[t1]){
      int nl=(t2+1)%3;
      if (dist[t3][nl]!=INF)continue;
      dist[t3][nl]=dist[t1][t2]+1;
      q.push(P(t3,nl));
    }
  }
  int ans=dist[t][0];
  if (ans==INF)ans=-1;
  else(ans=ans/3);
  cout<<ans;
}

### Prompt

Generate 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 (int i = 0; i < (n); ++i)
using namespace std;
typedef pair<int,int> P;

const int MOD=1000000007;
const int INF=1001001001;


int main(){
  int n,m;
  cin>>n>>m;
  vector<vector<int>> to(n);
  rep(i,m){
    int u,v;

    cin>>u>>v;
    u--;v--;
    to[u].push_back(v);
  }
  int s,t;
  cin>>s>>t;
  s--;t--;
  int dist[n][3];
  rep(i,n)rep(j,3)dist[i][j]=INF;
  dist[s][0]=0;
  queue<P> q;
  q.push(P(s,0));
  while (!q.empty()){
    int t1,t2;
    t1=q.front().first;
    t2=q.front().second;
    q.pop();
    for(int t3:to[t1]){
      int nl=(t2+1)%3;
      if (dist[t3][nl]!=INF)continue;
      dist[t3][nl]=dist[t1][t2]+1;
      q.push(P(t3,nl));
    }
  }
  int ans=dist[t][0];
  if (ans==INF)ans=-1;
  else(ans=ans/3);
  cout<<ans;
}

```