#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
#include <queue>
using namespace std;

#define pb push_back

const int N = 4e5 + 10;

int n, m, s, t;
long long d[N];
vector<int> G[N]; //g, g + n, g + 2*n

void bfs(int s, int t) {
	fill(d + 1, d + 3 * n + 1, -1);
	d[s] = 0; queue<int> q; q.push(s);
	while(q.size()) {
		int u = q.front(); q.pop();
		for(int &v : G[u]) if(d[v] == -1) {
			d[v] = d[u] + 1; q.push(v);
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; i ++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v + n);
		G[u + n].pb(v + 2 * n);
		G[u + 2 * n].pb(v);
	}
	scanf("%d%d", &s, &t);
	bfs(s, t);
	if(d[t] == -1) puts("-1");
	else printf("%d\n", d[t] / 3);
	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2

### Response

```cpp
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstdio>
#include <queue>
using namespace std;

#define pb push_back

const int N = 4e5 + 10;

int n, m, s, t;
long long d[N];
vector<int> G[N]; //g, g + n, g + 2*n

void bfs(int s, int t) {
	fill(d + 1, d + 3 * n + 1, -1);
	d[s] = 0; queue<int> q; q.push(s);
	while(q.size()) {
		int u = q.front(); q.pop();
		for(int &v : G[u]) if(d[v] == -1) {
			d[v] = d[u] + 1; q.push(v);
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; i ++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].pb(v + n);
		G[u + n].pb(v + 2 * n);
		G[u + 2 * n].pb(v);
	}
	scanf("%d%d", &s, &t);
	bfs(s, t);
	if(d[t] == -1) puts("-1");
	else printf("%d\n", d[t] / 3);
	return 0;
}

```