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 <iostream>
#include <algorithm>
#include <complex>
#include <utility>
#include <vector>
#include <string>
#include <queue>
#include <tuple>
#include <cmath>
#include <bitset>
#include <cctype>
#include <set>
#include <map>
#include <numeric>
#include <functional>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define PRINT(V) cout << V << "\n"
#define SORT(V) sort((V).begin(),(V).end())
#define RSORT(V) sort((V).rbegin(), (V).rend())
using namespace std;
using ll = long long;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
inline void Yes(bool condition){ if(condition) PRINT("Yes"); else PRINT("No"); }
template<class itr> void cins(itr first,itr last){
    for (auto i = first;i != last;i++){
        cin >> (*i);
    }
}
template<class itr> void array_output(itr start,itr goal){
    string ans = "",k = " ";
    for (auto i = start;i != goal;i++) ans += to_string(*i)+k;
    if (!ans.empty()) ans.pop_back();
    PRINT(ans);
}
ll gcd(ll a, ll b) {
    return a ? gcd(b%a,a) : b;
}
const ll INF = 1e18;
const ll MOD = 1000000007;
typedef pair<ll,ll> P;
const ll MAX = 20000;
constexpr ll nx[8] = {1,0,-1,0,-1,-1,1,1};
constexpr ll ny[8] = {0,1,0,-1,-1,1,-1,1};
vector<vector<bool>> done(100005,vector<bool>(3,0));
vector<vector<ll>> g(100005),d(100005,vector<ll>(3,INF));
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,m,u,v,s,t;
    cin >> n >> m;
    rep(i,m){
        cin >> u >> v;
        --u;
        --v;
        g[u].push_back(v);
    }
    cin >> s >> t;
    --s;
    --t;
    d[s][0] = 0;
    queue<P> q;
    q.push(P(s,0));
    while(!q.empty()){
        ll v,c;
        tie(v,c) = q.front();
        q.pop();
        done[v][c] = true;
        ll k = (c+1)%3;
        for(ll u:g[v]){
            if (done[u][k]) continue;
            chmin(d[u][k],d[v][c]+1);
            q.push(P(u,k));
        }
    }
    if (done[t][0]) PRINT(d[t][0]/3);
    else PRINT(-1);
}