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;
const int maxn=1e5+10,inf=1e9;
int dp[maxn][3],n,m;
bool vis[maxn][3];
vector<int> a[maxn];
struct pa{
    int val,nod,l;
}p;
int main(){
    int u,v;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u,&v);
        a[u].push_back(v);
    }
    for(int i=1;i<=n;i++)
        for(int j=0;j<3;j++)
        dp[i][j]=inf;
    scanf("%d%d",&u,&v);
    dp[u][0]=0;
    queue<pa> q;
    q.push({0,u,0});
    while(!q.empty()){
        p=q.front();q.pop();
        if(vis[p.nod][p.l])continue;
        vis[p.nod][p.l]=true;
        for(int i:a[p.nod]){
            if(!vis[i][(p.l+1)%3]){
                dp[i][(p.l+1)%3]=p.val+1;
                q.push({p.val+1,i,(p.l+1)%3});
            }
        }
    }
    printf("%d\n",dp[v][0]==inf?-1:dp[v][0]/3);
}
