#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef pair<int,int> P;
int N,M,S,T;
int dp[100010][3] = {};
int inf = 1e9;
vector<vector<int>> v(100010);

int main(){
    cin >> N >> M;
    int a,b;
    for(int i=0;i<M;i++){
        cin >> a >> b;
        v[a].push_back(b);
    }
    cin >> S >> T;
    for(int i=1;i<=N;i++) for(int j=0;j<=2;j++) dp[i][j] = inf;
    dp[S][0] = 0;
    queue<P> Q;
    Q.push({S,0});
    while(!Q.empty()){
        P p = Q.front(); Q.pop();
        for(auto x:v[p.first]){
            int ne = (p.second+1)%3;
            if(dp[x][ne]==inf){
                dp[x][ne] = dp[p.first][p.second]+1;
                Q.push({x,ne});
            }
        }
    }
    cout << (dp[T][0]!=inf? dp[T][0]/3:-1) << endl;
}
### Prompt

Your challenge is to write 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 <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef pair<int,int> P;
int N,M,S,T;
int dp[100010][3] = {};
int inf = 1e9;
vector<vector<int>> v(100010);

int main(){
    cin >> N >> M;
    int a,b;
    for(int i=0;i<M;i++){
        cin >> a >> b;
        v[a].push_back(b);
    }
    cin >> S >> T;
    for(int i=1;i<=N;i++) for(int j=0;j<=2;j++) dp[i][j] = inf;
    dp[S][0] = 0;
    queue<P> Q;
    Q.push({S,0});
    while(!Q.empty()){
        P p = Q.front(); Q.pop();
        for(auto x:v[p.first]){
            int ne = (p.second+1)%3;
            if(dp[x][ne]==inf){
                dp[x][ne] = dp[p.first][p.second]+1;
                Q.push({x,ne});
            }
        }
    }
    cout << (dp[T][0]!=inf? dp[T][0]/3:-1) << endl;
}
```