// Slower C++ solution:
//#pragma GCC target("avx2")

//#pragma GCC optimize("O3")

   

//#include <x86intrin.h>

#include <bits/stdc++.h>   

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

 

using namespace __gnu_pbds;

using namespace std;



template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

     

#define F first

#define S second           

#define lb lower_bound               

#define ub upper_bound

#define pb push_back

#define pf push_front    

#define ppb pop_back

#define mp make_pair                                     

#define bpp __builtin_popcountll                                                                                        

#define sqr(x) ((x) * (x)) 

#define al 0x3F3F3F3F

#define sz(x) (int)x.size()

#define all(x) x.begin(), x.end()

#define in insert

#define ppf pop_front

#define endl '\n'

//#define int long long

 

typedef unsigned long long ull;

typedef long long ll;

typedef long double ld;

typedef pair <int, int> pii;

 

const int mod = (int)1e9 + 7;

const int N = (int)3e5 + 123;

const ll inf = (ll)1e18 + 1;



const double pi = acos (-1.0);

const double eps = 1e-7;                  

const int dx[] = {0, 0, 1, 0, -1};

const int dy[] = {0, 1, 0, -1, 0};



int n, m, d[N][3], s, t;

queue <int> q;

vector <int> g[N];



inline void boost () {                    

   ios_base :: sync_with_stdio (NULL);

   cin.tie (NULL), cout.tie (NULL);          

}                                                     



inline void Solve () {

   cin >> n >> m;

   for (int i = 1; i <= m; i ++) {

      int x, y;

      cin >> x >> y;

      g[x].pb (y);

   }

   for (int i = 1; i <= n; i ++)

      for (int j = 0; j < 3; j ++)

         d[i][j] = mod;

   cin >> s >> t;

   d[s][0] = 0;

   q.push (s);

   while (sz (q)) {

      int v = q.front ();

      q.pop ();

      for (int ost = 0; ost < 3; ost ++) {

         int nxt = (ost + 1) % 3;

         for (auto to : g[v]) {    

            if (d[to][nxt] > d[v][ost] + 1) {

               d[to][nxt] = d[v][ost] + 1;

               q.push (to);

            }

         }

      }

   }

   if (d[t][0] == mod) cout << -1, exit (0);

   cout << d[t][0] / 3;

}



main () {                                       

// freopen (".in", "r", stdin);    

// freopen (".out", "w", stdout);              

   boost ();

   int tt = 1;

   //cin >> tt;   

   while (tt --) {

      Solve ();

   }                                               

   return 0;      

}

                          

// Faster C++ solution for the same problem (measured speedup: 2.6475550488251955x):
#include<bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10;

vector<int> edg[N];

int dis[N][3];

int main() {

  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

  int n, m; cin >> n >> m;

  for(int i = 0; i < m; i++) {

    int u, v; cin >> u >> v;

    edg[u].push_back(v);

  }

  for(int i = 1; i <= n; i++) for(int j = 0; j < 3; j++) dis[i][j] = -3; // 设成-3是因为如果结果要除以3，不存在时-3/3=-1

  int s, t; cin >> s >> t;

  queue<pair<int, int>> q;

  dis[s][0] = 0;

  q.push({s, 0});

  while(!q.empty()) {

    const int cx = q.front().first, cy = q.front().second, ny = (cy + 1) % 3;

    q.pop();

    for(int i = 0; i < int(edg[cx].size()); i++) {

      const int nx = edg[cx][i];

      if(dis[nx][ny] == -3) {

        q.push({nx, ny});

        dis[nx][ny] = dis[cx][cy] + 1;

      }

    }

  }

  cout << dis[t][0] / 3 << "\n";

}