// 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: 2.664844789633333x):
#include <bits/stdc++.h>



#define MAX 100005



using namespace std;



int n, m, s, t;

vector<int> G[MAX];

int8_t mark[MAX];;



int bfs()

{

        queue<pair<int,int>> q;

        int ls = 0;

        int kkp = 0;

        q.emplace(s,0);



        while (!q.empty()) {

                auto curr = q.front();

                q.pop();

                if (curr.second != 3) {

                        kkp += (ls == 3);

                        ls = curr.second;



                        if ((mark[curr.first] & (1<<curr.second))) continue;

                        mark[curr.first] |= (1<<curr.second);

                        for (auto e : G[curr.first])

                                q.emplace(e, curr.second+1);

                } else {

                        if (curr.first == t) return kkp+1;

                        else if ((mark[curr.first] & (1<<curr.second))) continue;



                        mark[curr.first] |= (1<<curr.second);

                        ls = curr.second;

                        for (auto e : G[curr.first])

                                q.emplace(e, 1);

                }

        }

        return -1;

}



int main()

{

        cin >> n >> m;

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

                int u, v;

                cin >> u >> v;

                G[u].push_back(v);

        }

        cin >> s >> t;

        cout << bfs() << endl;

        return 0;

}