#include "bits/stdc++.h"
using namespace std;

const int N=1e5+20;

int n,m,u,v,dp[N][3],source,dest;
queue <pair <int,int>> q;
bool vis[N][3];
vector <int> adj[N];

void add(int node,int mod,int dist)
{
    q.push({node,mod});
    dp[node][mod]=dist;
    vis[node][mod]=true;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&u,&v);
        adj[u].push_back(v);
    }
    scanf("%d%d",&source,&dest);
    
    add(source,0,0);
    while(not q.empty())
    {
        auto p=q.front();
        q.pop();
        
        int node=p.first,mod=p.second;
        for(auto child:adj[node])
            if(not vis[child][(mod+1)%3])
                add(child,(mod+1)%3,dp[node][mod]+1);
    }
    
    if(not vis[dest][0]) printf("-1");
    else printf("%d",dp[dest][0]/3);
}
### Prompt

Construct a CPP code solution to the problem outlined:

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"
using namespace std;

const int N=1e5+20;

int n,m,u,v,dp[N][3],source,dest;
queue <pair <int,int>> q;
bool vis[N][3];
vector <int> adj[N];

void add(int node,int mod,int dist)
{
    q.push({node,mod});
    dp[node][mod]=dist;
    vis[node][mod]=true;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&u,&v);
        adj[u].push_back(v);
    }
    scanf("%d%d",&source,&dest);
    
    add(source,0,0);
    while(not q.empty())
    {
        auto p=q.front();
        q.pop();
        
        int node=p.first,mod=p.second;
        for(auto child:adj[node])
            if(not vis[child][(mod+1)%3])
                add(child,(mod+1)%3,dp[node][mod]+1);
    }
    
    if(not vis[dest][0]) printf("-1");
    else printf("%d",dp[dest][0]/3);
}
```