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 all(v) (v).begin(),(v).end()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define ri(x) scanf("%d",&(x))
#define ri2(x,y) scanf("%d %d",&(x),&(y))
#define ri3(x,y,z) scanf("%d %d %d",&(x),&(y),&(z))
#define rll(x) scanf("%lld",&(x))
#define rll2(x,y) scanf("%lld %lld",&(x),&(y))
#define rll3(x,y,z) scanf("%lld %lld %lld",&(x),&(y),&(z))
#define gc(x) ((x) = getchar())
using namespace::std;

const long double PI = acos(-1);
const long long MOD = 1000000000 +7;

typedef long long ll;
typedef pair<ll,ll> pll;
typedef pair<ll,pll> tll;
typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<tll> vtll;
typedef vector<string> vs;
typedef set<int> si;
typedef set<ii> sii;
typedef set<iii> siii;

ll gcd(ll a, ll b){ return b==0?a:gcd(b,a%b);}

ll add(ll a, ll b, ll m = MOD){
	if(a >= m) a %= m;
	if(b >= m) b %= m;
	if(a < 0) a += m;
	if(b < 0) b += m;
	ll res = a+b;
	if(res >= m or res <= -m) res %= m;
	if(res < 0) res += m;
	return res;
}

ll mul(ll a, ll b, ll m = MOD){
	if(a >= m) a %= m;
	if(b >= m) b %= m;
	if(a < 0) a += m;
	if(b < 0) b += m;
	ll res = a*b;
	if(res >= m or res <= -m) res %= m;
	if(res < 0) res += m;
	return res;
}

ll pow_mod(ll a, ll b, ll m = MOD){
	ll res = 1LL;
	a = a%m;
	while(b){
		if(b&1) res = mul(res,a,m);
		b >>= 1;
		a = mul(a,a,m);
	}
	return res;
}

ll fastexp(ll a, ll b){
	ll res = 1LL;
	while(b){
		if(b&1) res = res*a;
		b >>= 1;
		a *= a;
	}
	return res;
}

int gcdExtendido(int a, int b, int *x, int *y){
	if(a == 0){
		*x = 0;
		*y = 1;
		return b;
	}
	int x1, y1;
	int gcd = gcdExtendido(b%a,a,&x1,&y1);
	
	*x = y1-(b/a)*x1;
	*y = x1;
	return gcd;
}

int modInverso(int a, int m){
	int x, y;
	int g = gcdExtendido(a,m,&x,&y);
	if(g!=1) return -1;
	else return (x%m + m)%m;
}

/****************************************
*************P*L*A*N*T*I*L*L*A************
*****************************************/

const int MAX = 200000 + 5;
const int N = 100000+5;

int n;
int m;
vi G[N];
int D[N][3];
int pf[MAX];

void init(){
	for(int i=2; i<MAX; i++) pf[i] = i;
	for(int i=2; i*i <= MAX; i++){
		if(pf[i] == i){
			for(int j=i*i; j<MAX; j+=i) pf[j] = i;
		}
	}
}

void BFS(int src){
	memset(D,-1,sizeof D);
	D[src][0] = 0;
	queue<int> Q;
	Q.emplace(src);
	Q.emplace(0);
	while(!Q.empty()){
		int u = Q.front(); Q.pop();
		int c = Q.front(); Q.pop();
		for(int v : G[u]){
			int nc = c^1;
			if(D[v][nc] == -1){
				D[v][nc] = D[u][c] + 1;
				Q.emplace(v);
				Q.emplace(nc);
			}
		}
	}
}

int solve(int snk){
	if(D[snk][0] == 2) return 2;
	if(D[snk][1] == -1) return -1;
	int ans = D[snk][1];
	while(pf[ans] != ans) ans += 2;
	return ans;
}

int main(){
	ri2(n,m);
	int s, t;
	ri2(s,t);
	init();
	for(int i=0; i<m; i++){
		int u, v;
		ri2(u,v);
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}
	BFS(s);
	int ans = solve(t);
	printf("%d\n",ans);
	return 0;
}