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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const int INF=1e9;

vector<int> e[300000];
int d[300000];

void bfs(int s){
    fill(d,d+300000,-3);
    d[s]=0;
    queue<int> q;
    q.push(s);
    while(q.size()){
        int v=q.front();
        q.pop();
        for(int n:e[v]){
            if(d[n]==-3){
                q.push(n);
                d[n]=d[v]+1;
            }
        }
    }
}

int main(){
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        --a;--b;
        e[a].push_back(n+b);
        e[n+a].push_back(n+n+b);
        e[n+n+a].push_back(b);
    }
    int s,t;
    cin>>s>>t;
    --s;--t;
    bfs(s);
    cout<<d[t]/3<<endl;
}
