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>
#define ll long long
#define ull unsigned long long
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define inf 0x3f3f3f3f
using namespace std;

const int maxn = 100050;

int n,m;

bool vis[maxn][4];
vector<int>G[maxn];

struct node{
    int v,p;
};

int bfs(int s,int t){
    queue<node>q;
    q.push(node{s,0});
    while(!q.empty()){
        node u=q.front();q.pop();
        if(vis[u.v][u.p%3])continue;
        vis[u.v][u.p%3]=1;
        if(u.v==t&&u.p%3==0)return u.p/3;
        for(auto v:G[u.v]){
            q.push(node{v,u.p+1});
        }
    }
    return -1;
}

int main(){
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int x,y;scanf("%d%d",&x,&y);
        G[x].push_back(y);
    }
    int s,t;
    scanf("%d%d",&s,&t);
    int ans=bfs(s,t);
    cout<<ans<<endl;
}
