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;
const int maxn = 1e6+7;
const int maxm = 2e6+7;
const ll mod = 1e9+7;
int n,m;
bool vis[maxn][3];
vector<int> g[maxn];
int d[maxn][3];
priority_queue<pair<int, int>> qu;
int main()
{
    memset(d, -1, sizeof(d));
    scanf("%d%d", &n, &m);
    for(int i=1;i<=m;i++){
        int a,b;scanf("%d%d",&a,&b);
        g[a].push_back(b);
    }
    int s,t;
    scanf("%d%d", &s, &t);
    qu.push(make_pair(0, s*3));
    int ans = -1;
    while(!qu.empty()){
        int dis = qu.top().first, u = qu.top().second/3, ins = qu.top().second%3;qu.pop();
        if(u == t && ins == 0){
            ans = dis;
            break;
        }
        vis[u][ins] = 1;
        for(int v : g[u]){
            int nins = ins+1;
            int ndis = dis;
            if(nins==3){
                nins=0;
                ndis++;
            }
            if(!vis[v][nins]){
                vis[v][nins]=1;
                qu.push(make_pair(ndis, v*3+nins));
            }
        }
    }
    printf("%d\n", ans);


    return 0;
}