#include<bits/stdc++.h>
#define inf 1e9
using namespace std;
const int maxn=1e5+10;
vector<int>s[maxn];
vector<int> dis(3*maxn,inf);
int main()
{
    int m,n;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int x,y;
        cin>>x>>y;
        x--,y--;
        s[x].push_back(y);
    }
    int st,ed;
    cin>>st>>ed;
    st--,ed--;
       dis[st]=0;
   queue<int>que;
    que.push(st);
    while(!que.empty()){
        int x=que.front();
        que.pop();
        int sum=x%n;

        for(int i=0;i<s[sum].size();i++){
           int d=dis[x]+1;
           int y=s[sum][i]+n*(d%3);
           if(dis[y]!=inf)
             continue;
           dis[y]=d;
           que.push(y);

        }
    }
   // for(int i=1;i<=n;i++)
  //      cout<<dis[i]<<endl;
printf("%d\n",dis[ed]==inf? -1:dis[ed]/3);

    return 0;
}

### Prompt

Please formulate a cpp solution 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<bits/stdc++.h>
#define inf 1e9
using namespace std;
const int maxn=1e5+10;
vector<int>s[maxn];
vector<int> dis(3*maxn,inf);
int main()
{
    int m,n;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int x,y;
        cin>>x>>y;
        x--,y--;
        s[x].push_back(y);
    }
    int st,ed;
    cin>>st>>ed;
    st--,ed--;
       dis[st]=0;
   queue<int>que;
    que.push(st);
    while(!que.empty()){
        int x=que.front();
        que.pop();
        int sum=x%n;

        for(int i=0;i<s[sum].size();i++){
           int d=dis[x]+1;
           int y=s[sum][i]+n*(d%3);
           if(dis[y]!=inf)
             continue;
           dis[y]=d;
           que.push(y);

        }
    }
   // for(int i=1;i<=n;i++)
  //      cout<<dis[i]<<endl;
printf("%d\n",dis[ed]==inf? -1:dis[ed]/3);

    return 0;
}

```