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 pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef pair<int,ii> iii;
typedef pair<int, iii> iiii;
typedef pair<double, ii> dii;
typedef long long ll;
typedef pair<ll, ii> lii;
const int INF = 1000000000;
const double EPS = 1e-9;
#define LSOne(S) (S & (-S))
int mods(int a, int b) { return (b + (a%b)) % b; }

int n, m, a, b, s, t;
ll dist[300005];
vector<vi> adjList;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
 
    cin >> n >> m;
    adjList.resize(3*n);

    for (int i = 0; i < 3*n; i++)
        dist[i] = 1e18;

    for (int i = 0; i < m; i++) {
        cin >> a >> b;
        a--; b--;
        adjList[a].push_back(b+n);
        adjList[a+n].push_back(b+2*n);
        adjList[a+2*n].push_back(b);
    }

    cin >> s >> t;
    s--; t--;

    queue<int> q;
    q.push(s);
    dist[0] = 0;

    bool reached = 0;
    while (!q.empty() && !reached) {
        int u = q.front(); q.pop();
        for (int i = 0; i < adjList[u].size() && !reached; i++) {
            int v = adjList[u][i];
            if (dist[v] == 1e18) {
                dist[v] = dist[u]+1;
                if (v == t) reached = 1;
                q.push(v);
            }
        }
    }

    if (dist[t] == 1e18) cout << "-1\n";
    else cout << dist[t]/3 << '\n';
 
    return 0;
}