#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
typedef pair<int,int> P;

const int INF = 1001001001;

const int MAXE = 100010;
int dist[MAXE][3];

signed main()
{
  int n,m,s,t;
  cin>>n>>m;
  vector<vector<int>> to(n);
  rep(i,m) {
    int a,b; cin>>a>>b;
    --a; --b;
    to[a].push_back(b);
    
  }
  cin>>s>>t;
  --s; --t;
  rep(i,n)rep(j,3) dist[i][j]=INF;
  queue<P> q;
  q.push(P(s,0));
  dist[s][0]=0;
  while(!q.empty()){
    int v=q.front().first;
    int l=q.front().second;
    q.pop();
    int nl=(l+1)%3;
    for(int u: to[v]){
      if(dist[u][nl] != INF) continue;
      dist[u][nl]=dist[v][l]+1;
      q.push(P(u,nl));
    }
  }
  printf("%d\n",dist[t][0]==INF?-1:dist[t][0]/3);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
using namespace std;

#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
typedef pair<int,int> P;

const int INF = 1001001001;

const int MAXE = 100010;
int dist[MAXE][3];

signed main()
{
  int n,m,s,t;
  cin>>n>>m;
  vector<vector<int>> to(n);
  rep(i,m) {
    int a,b; cin>>a>>b;
    --a; --b;
    to[a].push_back(b);
    
  }
  cin>>s>>t;
  --s; --t;
  rep(i,n)rep(j,3) dist[i][j]=INF;
  queue<P> q;
  q.push(P(s,0));
  dist[s][0]=0;
  while(!q.empty()){
    int v=q.front().first;
    int l=q.front().second;
    q.pop();
    int nl=(l+1)%3;
    for(int u: to[v]){
      if(dist[u][nl] != INF) continue;
      dist[u][nl]=dist[v][l]+1;
      q.push(P(u,nl));
    }
  }
  printf("%d\n",dist[t][0]==INF?-1:dist[t][0]/3);
}

```