Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h> // ver2.3.8
#include <boost/math/tools/minima.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#define fmin brent_find_minima
#define int long long
#define endl "\n"
#define ALL(v) (v).begin(),(v).end()
#define COUNT(a,k) upper_bound(ALL(a),k)-lower_bound(ALL(a),k)
#define BIGGER(a,k) a.end()-upper_bound(ALL(a),k)
#define SMALLER(a,k) lower_bound(ALL(a),k)-a.begin()
#define Vi vector<int>
#define VVi vector<Vi>
#define Vs vector<string>
#define Pii pair<int,int>
#define VPii vector<Pii>
#define Tiii tuple<int,int,int>
#define VTiii vector<Tiii>
#define PQi priority_queue<int>
#define PQir priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define itos to_string
#define stoi stoll
#define FI first
#define SE second
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define cyes cout<<"yes"<<endl
#define cno cout<<"no"<<endl
#define sortr(v) sort(v,greater<>())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define leng(n) (int)(log10(n)+1)
#define dem(a,b) ((a+b-1)/(b))
#define Vin(a) rep(iI,0,a.size())cin>>a[iI]
#define INF 3000000000000000000  //  3.0*10^18(MAXの1/3くらい)
#define MAX LLONG_MAX
#define PI 3.141592653589793238462
#define MOD 1000000007  //  10^9 + 7

using namespace std;
/* fmin(f,L,R,100) で関数の最小値を(x,min)で返す　*/
using boost::math::tools::brent_find_minima;
/* cpp_int (任意の長さのint) */
using boost::multiprecision::cpp_int;

/* デバッグ用 */
void Vout(auto a){if(a.size()==0) cout<<"."<<endl;
else{rep(i,0,a.size()-1)cout<<a[i]<<' ';if(a.size()>0)cout<<a.back()<<endl;}return;}
void Verr(auto a){if(a.size()==0) cerr<<"."<<endl;
else{rep(i,0,a.size()-1)cerr<<a[i]<<' ';if(a.size()>0)cerr<<a.back()<<endl;}return;}
void VVout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())Vout(a[i]);}return;}
void VVerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())Verr(a[i]);}return;}
void VPout(auto a){if(a.size()==0)cout<<"."<<endl;
else{rep(i,0,a.size())cout<<a[i].FI<<' '<<a[i].SE<<endl;}return;}
void VPerr(auto a){if(a.size()==0)cerr<<"."<<endl;
else{rep(i,0,a.size())cerr<<a[i].FI<<' '<<a[i].SE<<endl;}return;}

/* 便利関数 */
int gcd(int a,int b){return b?gcd(b,a%b):a;} // 最大公約数gcd
int lcm(int a,int b){return a/gcd(a,b)*b;} // 最小公倍数lcm
int mypow(int x, int n, int m){ // 累乗x^n(mod m) O(log(n))
if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int scomb(int n, int r){if((n-r)<r)r=n-r; // nCr(小さい場合)
int a=1;for(int i=n;i>n-r;--i){a=a*i;}for(int i=1;i<r+1;++i){a=a/i;}return a;}
int comb(int n, int r){if((n-r)<r)r=n-r; // nCr(%MOD)
int a=1;for(int i=n;i>n-r;--i){a=a*i%MOD;}for(int i=1;
i<r+1;++i){a=a*mypow(i,MOD-2,MOD)%MOD;}return a%MOD;}
Vi stpowV(){Vi a(100001);
a[0]=1;repeq(i,1,100000)a[i]=a[i-1]*i%MOD;return a;}
//Vi stpow = stpowV();  // 階乗配列(%MOD)
Vi Vsum(Vi &v){Vi S(v.size()+1);rep(i,1,S.size())S[i]+=v[i-1]+S[i-1];return S;}// 累積和
int digit(int k,int i){string s = itos(k);return s[s.size()-i]-'0';} // i桁目の数字
void press(auto &v){v.erase(unique(ALL(v)),v.end());} // 圧縮
Vi zip(Vi b){int Z=b.size(); // 座標圧縮
Pii p[Z+10];int a=b.size();Vi l(a);for(int i=0;i<a;i++)
p[i]=mp(b[i],i);sort(p,p+a);int w=0;for(int i=0;i<a;i++)
{if(i&&p[i].first!=p[i-1].first)w++;l[p[i].second]=w;}return l;}
Vi beki2V(){Vi a(61);int q = 1;rep(i,0,61){a[i]=q;q*=2;}return a;}
//Vi beki2 = beki2V(); // 2^n配列
int modiv(int a,int b){return a*mypow(b,MOD-2,MOD)%MOD;} // a÷b(MOD環境) 

int n,m;
VVi from,to;
int S,G;
vector<VPii> newgr;


vector<vector<Pii>> gr;
Vi dist;
vector<int> prever;

void shortest(int s){
  int n = dist.size();
  prever = vector<int>(n,-1);
  rep(i,0,n) dist[i]=INF;
  dist[s] = 0;
  
  priority_queue<Tiii,vector<Tiii>,greater<Tiii>> q;
  q.push(make_tuple(0,s,-1));
  while(!q.empty()){
    int nowdist,now,past;
    tie(nowdist,now,past)=q.top();
    q.pop();
    if(nowdist>dist[now]) continue;
    dist[now]=nowdist;
    rep(i,0,gr[now].size()){
      int next=gr[now][i].first,cost=gr[now][i].second;
      if(next!=past && nowdist+cost<dist[next]){
        prever[next] = now; // テスト
        q.push(make_tuple(nowdist+cost,next,now));
        dist[next]=nowdist+cost;
      }
    }
  }
  return;
}

vector<int> get_path(int t){ //頂点tへの最短路
  vector<int> path;
  for(;t!=-1;t=prever[t]){
    path.push_back(t);
  }
  reverse(path.begin(), path.end());
  return path;
}


signed main() {
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(12);
  cin >> n >> m;
  
  Vi x(m),y(m);
  
  from = VVi(n);
  to = VVi(n);
  
  int qw,er;
  rep(i,0,m){
    cin >> qw >> er;
    qw--,er--;
    x[i] = qw;
    y[i] = er;
    to[qw].pb(er);
    from[er].pb(qw);
  }
  
  newgr = vector<VPii>(n);
  set<Pii> kind;
  
  rep(i,0,m){
    rep(j,0,from[x[i]].size()){
      rep(k,0,to[y[i]].size()){
        qw = from[x[i]][j];
        er = to[y[i]][k];
        if(kind.count(mp(qw,er)) == 1) continue;
        kind.insert(mp(qw,er));
        newgr[qw].pb(mp(er,1));
      }
    }
  }
  
  cin >> S >> G;
  S--,G--;
  
  gr = newgr;
  dist = Vi(n);
  
  shortest(S);
  
  if(dist[G] >= INF/3) cout << -1 << endl;
  else cout << dist[G] << endl;
  
  return 0;
}