// 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: 1.4676972182267232x):
#include <bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef pair<ll,ll> pi;

typedef vector <ll> vi;

typedef vector <pi> vpi;

#define f first

#define s second

#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)

#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)

#define pb push_back

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

#define lbd(x, y) lower_bound(all(x), y)

#define ubd(x, y) upper_bound(all(x), y)

#define aFOR(i,x) for (auto i: x)

#define mem(x,i) memset(x,i,sizeof x)

#define fast ios_base::sync_with_stdio(false),cin.tie(0)

#define maxn 100001



int N,M,S,T;

vi adj[maxn];

vpi adj2[maxn];

bool vis[maxn][3];

int dist[maxn][3];

vpi nodes[3];



int main(){

	fast;

	

	cin>>N>>M;

	

	FOR(i,0,M-1){

		int a,b; cin>>a>>b; adj[a].pb(b);

	}

	

	cin>>S>>T;

	



		

	queue <pi> q;

	q.push(pi(S,0));

		

	while (!q.empty()){

		pi cur = q.front(); q.pop();

			

		if (vis[cur.f][cur.s%3]) continue;

		vis[cur.f][cur.s%3] = 1;

		

		dist[cur.f][cur.s%3] = cur.s;

		aFOR(i,adj[cur.f]){

			q.push(pi(i,cur.s+1));

		}

	}

	

	if (vis[T][0] == 1) cout<<dist[T][0]/3;

	else cout<<-1;



		

}