#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

typedef pair<int,int> PII;
const int M=1e5+5;
vector<int> g[M];
int d[M][3];
queue<PII> qu;

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	
	int u,v;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	memset(d,-1,sizeof(d));
	qu.emplace(s,0);
	d[s][0]=0;
	while(!qu.empty())
	{
		PII node=qu.front(); qu.pop();
		int u=node.first,step=node.second;
		if(u==t&&step==0)
		{
			printf("%d\n",d[u][0]/3);
			return 0;
		}
		int nstep=(step+1)%3;
		for(int v:g[u])
		{
			if(d[v][nstep]!=-1) continue;
			qu.emplace(v,nstep);
			d[v][nstep]=d[u][step]+1;
		}
	}
	printf("-1\n");
	return 0;
}
### Prompt

In CPP, your task is to solve 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 <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

typedef pair<int,int> PII;
const int M=1e5+5;
vector<int> g[M];
int d[M][3];
queue<PII> qu;

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	
	int u,v;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
	}
	int s,t;
	scanf("%d%d",&s,&t);
	memset(d,-1,sizeof(d));
	qu.emplace(s,0);
	d[s][0]=0;
	while(!qu.empty())
	{
		PII node=qu.front(); qu.pop();
		int u=node.first,step=node.second;
		if(u==t&&step==0)
		{
			printf("%d\n",d[u][0]/3);
			return 0;
		}
		int nstep=(step+1)%3;
		for(int v:g[u])
		{
			if(d[v][nstep]!=-1) continue;
			qu.emplace(v,nstep);
			d[v][nstep]=d[u][step]+1;
		}
	}
	printf("-1\n");
	return 0;
}
```