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;
#define ll long long
#define pb push_back
#define sz(x) (int)x.size()
typedef pair<int,int>pii;
const int mxN=1e5+10,M=1e9+7;
int n,m,a,b;
int s,t;
int vis[mxN][4];
vector<int>adj[mxN];

int bfs(){
    queue<pii>fila;
    fila.push({s,0});    
    while(!fila.empty()){
        int u=fila.front().first;
        int d=fila.front().second;
        fila.pop();
        if(vis[u][d%3])continue;
        vis[u][d%3]=true;
        if(u==t && d%3==0)return d/3;
        for(int v:adj[u]){
            fila.push({v,d+1});
        }
    }
    return -1;
}



// LONG LONG CASE
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>a>>b;
        adj[a].pb(b);
    }
    cin>>s>>t;    
    cout<< bfs();


}