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

const int Maxn = 100005;
int n, m, st, ed;
int dist[Maxn][3];
vector <int> g[Maxn];

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; ++ i) {
		int u, v;
		scanf("%d %d", &u, &v);
		g[u].push_back(v);
	}
	scanf("%d %d", &st, &ed);
	queue <pair <int, int> > que;
	que.push({st, 0});
	memset(dist, 0x3f, sizeof (dist));
	dist[st][0] = 0;
	while (!que.empty()) {
		int u = que.front().first;
		int d = que.front().second;
		que.pop();
		for (auto &v: g[u])
			if (dist[v][(d + 1) % 3] > dist[u][d] + 1) {
				dist[v][(d + 1) % 3] = dist[u][d] + 1;
				que.push({v, (d + 1) % 3});
			}
	}
	if (dist[ed][0] < 0x3f3f3f3f)
		printf("%d\n", dist[ed][0] / 3);
	else
		puts("-1");
	return 0;
}
### Prompt

Please create a solution in Cpp to the following problem:

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

const int Maxn = 100005;
int n, m, st, ed;
int dist[Maxn][3];
vector <int> g[Maxn];

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < m; ++ i) {
		int u, v;
		scanf("%d %d", &u, &v);
		g[u].push_back(v);
	}
	scanf("%d %d", &st, &ed);
	queue <pair <int, int> > que;
	que.push({st, 0});
	memset(dist, 0x3f, sizeof (dist));
	dist[st][0] = 0;
	while (!que.empty()) {
		int u = que.front().first;
		int d = que.front().second;
		que.pop();
		for (auto &v: g[u])
			if (dist[v][(d + 1) % 3] > dist[u][d] + 1) {
				dist[v][(d + 1) % 3] = dist[u][d] + 1;
				que.push({v, (d + 1) % 3});
			}
	}
	if (dist[ed][0] < 0x3f3f3f3f)
		printf("%d\n", dist[ed][0] / 3);
	else
		puts("-1");
	return 0;
}
```