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: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <utility>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include <climits>
#include <numeric>
#include<stdio.h>
#include <queue>
#include <iomanip>
#include <float.h>
#include <set>
#include<functional>
#include <stack>
#include <time.h>
#include <climits>
#include <bitset>
#include <fstream>
using namespace std;
typedef pair<int, int>p;
long long gcd(long long a, long long b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
vector<long long> divisor(long long n) {
	vector< long long> ret;
	for (long long i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(begin(ret), end(ret));
	return (ret);
}
struct LazySegmentTree {
private:
	int n;
	vector<int> node, lazy;
	vector<bool> lazyFlag;

public:
	LazySegmentTree(vector<int> v) {
		int sz = (int)v.size();
		n = 1; while (n < sz) n *= 2;
		node.resize(2 * n - 1);
		lazy.resize(2 * n - 1, INT_MAX);
		lazyFlag.resize(2 * n - 1, false);

		for (int i = 0; i<sz; i++) node[i + n - 1] = v[i];
		for (int i = n - 2; i >= 0; i--) node[i] = min(node[i * 2 + 1], node[i * 2 + 2]);
	}

	void lazyEvaluate(int k, int l, int r) {
		if (lazyFlag[k]) {
			node[k] = lazy[k];
			if (r - l > 1) {
				lazy[k * 2 + 1] = lazy[k * 2 + 2] = lazy[k];
				lazyFlag[k * 2 + 1] = lazyFlag[k * 2 + 2] = true;
			}
			lazyFlag[k] = false;
		}
	}

	void update(int a, int b, int x, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		lazyEvaluate(k, l, r);
		if (b <= l || r <= a) return;
		if (a <= l && r <= b) {
			lazy[k] = x;
			lazyFlag[k] = true;
			lazyEvaluate(k, l, r);
		}
		else {
			update(a, b, x, 2 * k + 1, l, (l + r) / 2);
			update(a, b, x, 2 * k + 2, (l + r) / 2, r);
			node[k] = min(node[2 * k + 1], node[2 * k + 2]);
		}
	}

	int find(int a, int b, int k = 0, int l = 0, int r = -1) {
		if (r < 0) r = n;
		lazyEvaluate(k, l, r);
		if (b <= l || r <= a) return INT_MAX;
		if (a <= l && r <= b) return node[k];
		int vl = find(a, b, 2 * k + 1, l, (l + r) / 2);
		int vr = find(a, b, 2 * k + 2, (l + r) / 2, r);
		return min(vl, vr);
	}
};
long long mod = 1000000007;
#define ll long long
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

					 // Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};
class BIT
{
public:
	vector<long long> bit;
	long long M;

	BIT(long long M) :
		bit(vector<long long>(M + 1, 0)), M(M) {}

	long long sum(long long i) {
		if (i == 0)return 0;
		if (!i) return 0;
		return (bit[i] + sum(i - (i&-i))) % mod;
	}

	void add(long long i, long long x) {
		if (i > M) return;
		bit[i] += x%mod;
		add(i + (i&-i), x);
	}
};
vector<int>graph[100004];
int n, m, s, t;
int dist[100004][3];
signed main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(16);
	cin >> n >> m;
	int u, v;
	for (int i = 0; i < m; i++) {
		cin >> u >> v;
		graph[u].push_back(v);
	}
	cin >> s >> t;
	queue<p>que;
	que.push(p(s, 0));
	for (int i = 0; i < 100004; i++) {
		for (int j = 0; j < 3; j++) {
			dist[i][j] = INT_MAX;
		}
	}
	dist[s][0] = 0;
	while (!que.empty()) {
		int now = que.front().first;
		int par = (que.front().second+1)%3;
		int d = dist[now][que.front().second];
		que.pop();
		for (int i = 0; i < graph[now].size(); i++) {
			int to = graph[now][i];
			if (dist[to][par]>d + 1) {
				dist[to][par] = d + 1;
				que.push(p(to, par));
			}
		}
	}
	if (dist[t][0] == INT_MAX)cout << -1;
	else cout << dist[t][0] / 3;
}