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 INF (1LL<<30)
#define DEKAI 1000000007
#define lp(i,n) for(int i=0;i<n;i++)
#define int long long
vector<int> g[100100];
int a[100100][3];

signed main(){
	int n,m;
	cin>>n>>m;
	lp(i,100100){
		lp(j,3)a[i][j]=INF;
	}
	lp(i,m){
		int a,b;
		cin>>a>>b;
		g[a].push_back(b);
	}
	int s,t;
	cin>>s>>t;
	queue<pair<int,int>> q;
	q.push({s,2});
	a[s][2]=0;
	int ans=-1;
	while(!q.empty()){
		int now=q.front().first;
		int cnt=q.front().second;
		q.pop();
		if(now==t&&cnt==2){
			ans=a[now][cnt];
			break;
		}
		int gx=a[now][cnt];
		gx++;
		cnt++;
		cnt%=3;
		lp(i,g[now].size()){
			int next=g[now][i];
			if(a[next][cnt]>gx){
				a[next][cnt]=gx;
				q.push({next,cnt});
			}
		}
	}
	if(ans==-1) cout<<-1<<endl;
	else cout<<ans/3<<endl;

	return 0;
}