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"

#pragma warning(disable:4996)
using namespace std;

vector<vector<int>>djk(const vector<vector<int>>&edges, const int S) {
	vector<vector<int>>dp(edges.size(),vector<int>(2,1e9));
	dp[S][0]=0;

	queue<pair<int,int>>que;
	que.emplace(S,0);
	while (!que.empty()) {
		auto p(que.front());
		que.pop();
		const int now(p.first);
		const bool odd(p.second);

		for (auto e : edges[now]) {
			const int nextcost(dp[now][odd]+1);
			if (dp[e][!odd] > nextcost) {
				dp[e][!odd]=nextcost;
				que.emplace(e,!odd);
			}
		}
	}
	return dp;
}
vector<long long int> divisor(long long int n) {
	vector<long long int> res;
	for (long long int i = 1; i * i <= n; ++i) {
		if (n % i == 0) {
			res.push_back(i);
			if (i * i != n) res.push_back(n / i);
		}
	}
	sort(begin(res), end(res));
	return res;
}

int main() {
	int N,M,S,T;cin>>N>>M>>S>>T;
	S--;T--;
	vector<vector<int>>edges(N);
	for (int i = 0; i < M; ++i) {
		int a,b;cin>>a>>b;
		a--;b--;
		edges[a].push_back(b);
		edges[b].push_back(a);
	}
	auto dp(djk(edges,S));
	int ans=dp[T][1];

	if (dp[T][0] == 2) {
		ans=2;
	}
	else {
		if(ans==1)ans=3;
		else if (ans > 1e8) {
			ans=-1;
		}
		else {
			while (divisor(ans).size()!=2) {
				ans+=2;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}