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 (GCC 9.2.1) (judge verdict: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <signal.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); bool s=false; if(c=='-'){s=true;c=getchar();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar(); int s=0; if(c=='-'){s=1;c=getchar();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar(); while(c==' '||c=='\n') c=getchar(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar(); } *s='\0'; return len; }

int edge[300000][2];

int qmem[600000];

int *e[300000];
int el[300000];

int d[300000];
int a[1000000][2];
int s=0;

void push(int x, int y){
	int c=s++, p=(c-1)/2;
	while(c>0 && x<a[p][0]){
		a[c][0]=a[p][0];
		a[c][1]=a[p][1];
		c=p, p=(c-1)/2;
	}
	a[c][0]=x;
	a[c][1]=y;
}
int pop(int *t){
	int td=a[0][0];
	*t=a[0][1];
	s--;
	int x=a[s][0], y=a[s][1];
	int p=0, c=p*2+1;
	while(c<s && a[c=(c+1<s&&a[c+1][0]<a[c][0])?c+1:c][0]<x){
		a[p][0]=a[c][0];
		a[p][1]=a[c][1];
		p=c, c=p*2+1;
	}
	a[p][0]=x;
	a[p][1]=y;
	return td;
}

void dike(int x){
	push(0,x);
	d[x]=0;
	while(s){
		int p;
		if(pop(&p)>d[p]) continue;
		for(int i=0; i<el[p]; i++){
			int to=e[p][i];
			long now=d[p]+1;
			if(now<d[to]){
				d[to]=now;
				push(now,to);
			}
		}
	}
}

int main(void){
	int n=nextint();
	int m=nextint();
	for(int i=0; i<m; i++){
		int A=nextint()-1;
		int B=nextint()-1;
		edge[i*3+0][0]=A*3+0; edge[i*3+0][1]=B*3+1; el[A*3+0]++;
		edge[i*3+1][0]=A*3+1; edge[i*3+1][1]=B*3+2; el[A*3+1]++;
		edge[i*3+2][0]=A*3+2; edge[i*3+2][1]=B*3+0; el[A*3+2]++;
	}
	int ne=0;
	for(int i=0; i<n*3; i++){
		e[i]=&qmem[ne];
		ne+=el[i];
		el[i]=0;
	}
	for(int i=0; i<m*3; i++){
		int A=edge[i][0];
		int B=edge[i][1];
		e[A][el[A]++]=B;
	}
	int s=(nextint()-1)*3;
	int t=(nextint()-1)*3;
	for(int i=0; i<n*3; i++){
		d[i]=INT_MAX;
	}
	dike(s);
	printf("%d\n", d[t]==INT_MAX?-1:d[t]/3);
}