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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
int u[100005], v[100005];
int asc(const void* x, const void* y)
{
	if (u[*(int*)x] > u[*(int*)y])return 1;
	if (u[*(int*)x] < u[*(int*)y])return -1;
	if (v[*(int*)x] > v[*(int*)y])return 1;
	if (v[*(int*)x] < v[*(int*)y])return -1;
	return 0;
}
int p[100005][3];
int h[1000006], l;
void f1(int n)
{
	h[l] = n;
	int now = l;
	l++;
	while (now > 0)
	{
		if (p[h[now] / 3][h[now] % 3] < p[h[(now - 1) / 2] / 3][h[(now - 1) / 2] % 3])
		{
			int f = h[now];
			h[now] = h[(now - 1) / 2];
			h[(now - 1) / 2] = f;
		}
		now = (now - 1) / 2;
	}
	return;
}
int f2()
{
	int res = h[0];
	l--;
	h[0] = h[l];
	int now = 0;
	for (;;)
	{
		if (2 * now + 2 < l)
		{
			if (p[h[2 * now + 1] / 3][h[2 * now + 1] % 3] > p[h[2 * now + 2] / 3][h[2 * now + 2] % 3])
			{
				if (p[h[2 * now + 2] / 3][h[2 * now + 2] % 3] < p[h[now] / 3][h[now] % 3])
				{
					int f = h[now];
					h[now] = h[2 * now + 2];
					h[2 * now + 2] = f;
				}
				now = 2 * now + 2;
			}
			else
			{
				if (p[h[2 * now + 1] / 3][h[2 * now + 1] % 3] < p[h[now] / 3][h[now] % 3])
				{
					int f = h[now];
					h[now] = h[2 * now + 1];
					h[2 * now + 1] = f;
				}
				now = 2 * now + 1;
			}
		}
		else if (2 * now + 1 < l)
		{
			if (p[h[2 * now + 1] / 3][h[2 * now + 1] % 3] < p[h[now] / 3][h[now] % 3])
			{
				int f = h[now];
				h[now] = h[2 * now + 1];
				h[2 * now + 1] = f;
			}
			now = 2 * now + 1;
		}
		else
			break;
	}
	return res;
}
int main()
{
	int n, m;
	scanf("%d %d", &n, &m);
	int i, j;
	for (i = 0; i < m; i++)
	{
		scanf("%d %d", &u[i], &v[i]);
		u[i]--;
		v[i]--;
	}
	int s, t;
	scanf("%d %d", &s, &t);
	s--;
	t--;
	if (m == 0)
	{
		printf("-1\n");
		return 0;
	}
	int c[100005];
	for (i = 0; i < m; i++)
		c[i] = i;
	qsort(c, m, sizeof(int), asc);
	int visited[100005][3];
	for (i = 0; i < n; i++)
		visited[i][0] = visited[i][1] = visited[i][2] = 0;
	for (i = 0; i < n; i++)
		p[i][0] = p[i][1] = p[i][2] = 1000000009;
	l = 0;
	f1(3 * s);
	p[s][0] = 0;
	int min, mid, max;
	int a, b;
	while (l > 0)
	{
		a = f2();
		b = a % 3;
		a /= 3;
		if (visited[a][b] > 0)
			continue;
		visited[a][b] = 1;
		min = 0;
		max = m;
		while (max - min > 1)
		{
			mid = (max + min) / 2;
			if (u[c[mid]] < a)
				min = mid;
			else
				max = mid;
		}
		if (u[c[min]] < a)
			min++;
		if (min < m)
		{
			while (u[c[min]] == a)
			{
				if (p[v[c[min]]][(b + 1) % 3] > p[a][b] + 1)
				{
					p[v[c[min]]][(b + 1) % 3] = p[a][b] + 1;
					f1(3 * v[c[min]] + (b + 1) % 3);
				}
				min++;
				if (min == m)
					break;
			}
		}
	}
	if (p[t][0] > 10000007)
		printf("-1\n");
	else
		printf("%d\n", p[t][0] / 3);
	return 0;
}