Problem: Fox Jiro is one of the staffs of the ACM-ICPC 2018 Asia Yokohama Regional Contest and is responsible for designing the network for the venue of the contest. His network consists of $N$ computers, which are connected by $M$ cables. The $i$-th cable connects the $a_i$-th computer and the $b_i$-th computer, and it carries data in both directions. Your team will use the $S$-th computer in the contest, and a judge server is the $T$-th computer.

He decided to adjust the routing algorithm of the network to maximize the performance of the contestants through the magical power of prime numbers. In this algorithm, a packet (a unit of data carried by the network) is sent from your computer to the judge server passing through the cables a prime number of times if possible. If it is impossible, the contestants cannot benefit by the magical power of prime numbers. To accomplish this target, a packet is allowed to pass through the same cable multiple times.

You decided to write a program to calculate the minimum number of times a packet from $S$ to $T$ needed to pass through the cables. If the number of times a packet passes through the cables cannot be a prime number, print $-1$.



Input

The input consists of a single test case, formatted as follows.


$N$ $M$ $S$ $T$
$a_1$ $b_1$
$\vdots$
$a_M$ $b_M$


The first line consists of four integers $N, M, S,$ and $T$ ($2 \leq N \leq 10^5, 1 \leq M \leq 10^5, 1 \leq S, T \leq N, S \ne T$). The $i$-th line of the following $M$ lines consists of two integers $a_i$ and $b_i$ ($1 \leq a_i < b_i \leq N$), which means the $i$-th cables connects the $a_i$-th computer and the $b_i$-th computer in the network. You can assume that the network satisfies the following conditions.

* The network has no multi-edge, i.e.,$(a_i, b_i) \ne (a_j, b_j)$ for all $i,j$ ($1 \leq i < j \leq M$).
* The packets from $N$ computers are reachable to $T$ by passing through some number of cables. The number is not necessarily a prime.

Output

If there are ways such that the number of times a packet sent from $S$ to $T$ passes through the cables is a prime number, print the minimum prime number of times in one line. Otherwise, print $-1$.

Examples

Input

5 5 1 5
1 2
1 4
2 3
3 4
3 5


Output

3


Input

5 4 1 5
1 2
2 3
3 4
4 5


Output

-1


Input

2 1 1 2
1 2


Output

3


Input

3 3 1 2
1 2
1 3
2 3


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<29;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-9;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

class Graph{
	private:
	int n;
	vvp g;
	public:
	int DIJ(int s,int t){
		priority_queue<P> q;
		vi d(n,inf);
		d[s]=0;
		q.push({0,s});
		while(!q.empty()){
			P p=q.top();
			q.pop();
			int v=p.second;
			if(d[v]<-p.first) continue;
			for(auto i:g[v]){
				int u=i.first,D=d[v]+i.second;
				if(d[u]>D){
					d[u]=D;
					q.push({-D,u});
				}
			}
		}
		return d[t];
	}
	Graph(int v){
		n=v;
		g=vvp(v);
	}
	void add_edge(int s,int t,int c){
		g[s].push_back({t,c});
		g[t].push_back({s,c});
	}
};

bool isPrime(ll n){
	if(n<=1) return 0;
	for(ll k=2;k*k<=n;k++) if(n%k==0) return 0;
	return 1;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,m,s,t;
	cin>>n>>m>>s>>t;
	Graph g(2*n);
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		g.add_edge(u,v+n,1);
		g.add_edge(u+n,v,1);
	}
	s--;t--;
	int x=g.DIJ(s,t);
	if(x==2) cout<<2<<endl;
	else{
		int y=g.DIJ(s,t+n);
		if(y==inf) cout<<-1<<endl;
		else{
			while(!isPrime(y)) y+=2;
			cout<<y<<endl;
		}
	}
}