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>
using namespace std;


int main() {
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--, t--;
  vector<vector<int>> graph(n, vector<int>());
  for (int i = 0; i < m; i++){
    int a, b;
    cin >> a >> b;
    a--, b--;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }

  vector<vector<int>> dists(n, vector<int>(2, 2147483647 / 2));
  dists[s][0] = 0;
  queue<pair<int, int>> q{};
  q.push(make_pair(s, 0));
  while (q.size() >= 1){
    auto elem = q.front();q.pop();
    auto parity = elem.second;
    auto ind = elem.first;
    auto dist = dists[ind][parity];
    for (auto&& item : graph[ind]){
      if (dists[item][parity ^ 1] <= dist + 1) continue;
      dists[item][parity ^ 1] = dist + 1;
      q.push(make_pair(item, parity ^ 1));
    }
  }
  if (dists[t][0] == 2){
    cout << 2 << endl;
    return 0;
  }
  if (dists[t][1] >= 10000000){
    cout << -1 << endl;
    return 0;
  }
  auto res = dists[t][1];
  while (true){
    if (res == 1) goto end;
    for (int i = 2; i * i <= res; i++){
      if (res % i == 0) goto end;
    }
    cout << res << endl;
    return 0;
  end:;
    res += 2;
  }
}