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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vector<int>> vvi;
typedef vector<string> vs;
typedef unordered_set<int> si;
typedef pair<int, int> pp;

#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#define all(e) (e).begin(), (e).end()
#define pb push_back


template <typename T>
void printV(vector<T> A) {
    if (A.empty()) cerr << "EMTPY\n";
    else { 
        cerr << "[";
        for (int i = 0; i < A.size() -1 ; ++i) {
            cerr << A[i] << ", ";
        }
        if (A.size()) cerr << A.back() << "]\n";
    }
}


template <typename T>
void printVV(vector<vector<T>> A) {
    cerr << "[";
    for (auto v : A) {
        printV<T>(v);
    }
    cerr << "]";
}

template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
    cerr << name << ": " << arg1 << endl;
}

template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&& ...args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << " |";
    __f(comma + 1, args...);
}

struct Info { 
    int node, ith, acc;
};

const int MAXN = 1e5 + 10;
int dp[MAXN][3];


int main(int argc, char **argv) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m; cin >> n >> m;
    auto graph = vvi(n);
    memset(dp,-1,sizeof(dp));
    for (int i = 0; i < m; ++i) {
        int u, v; cin >> u >> v;
        u--, v--;
        graph[u].pb(v);
    }
    int s, t; cin >> s >> t;
    s--, t--;
    queue<Info> q;
    q.push({s,0,0});
    dp[s][0] = 0;
    while (!q.empty()) {
        Info& i  = q.front(); q.pop();
        int node = i.node, ith = i.ith, acc = i.acc;
        if (node == t && ith == 0) {
            cout << acc << endl;
            return 0;
        }
        for (auto neigh: graph[node]) {
            int nith = (ith+1) % 3, nacc = acc + (ith == 0);
            if (dp[neigh][nith] == -1) {
                dp[neigh][nith] = nacc;
                q.push({neigh, nith, nacc});
            }
        }
    }
    cout << -1 << endl;



    return 0;
}