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 For(i, a, b) for(int (i)=(int)(a); (i)<(int)(b); ++(i))
#define rFor(i, a, b) for(int (i)=(int)(a)-1; (i)>=(int)(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;
template<class T> bool chmax(T &a, const T &b){if(a<b){a=b; return true;} return false;}
template<class T> bool chmin(T &a, const T &b){if(a>b){a=b; return true;} return false;}
template<class T> T div_floor(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>=0 ? a/b : (a+1)/b-1;
}
template<class T> T div_ceil(T a, T b){
    if(b < 0) a *= -1, b *= -1;
    return a>0 ? (a-1)/b+1 : a/b;
}

constexpr lint mod = 1e9+7;
constexpr lint INF = mod * mod;
constexpr int MAX = 200010;

vector<int> min_factor, prime;

void sieve(int n){
    min_factor.resize(n+1, 0);
    For(i, 2, n+1){
        if(min_factor[i] == 0){
            min_factor[i] = i;
            prime.push_back(i);
        }
        for(int x: prime){
            if(x*i > n || x > i) break;
            min_factor[x*i] = x;
        }
    }
}

bool is_prime(int x){
    return x > 1 && min_factor[x] == x;
}

vector<pii> prime_f(int n){
    vector<pii> ret;
    while(n > 1){
        if(ret.empty() || ret.rbegin()->fi != min_factor[n]){
            ret.emplace_back(min_factor[n], 1);
        }
        else ++ret.rbegin()->se;
        n /= min_factor[n];
    }
    return ret;
}

int main(){
    int n, m, s, t;
    scanf("%d%d%d%d", &n, &m, &s, &t);
    --s; --t;
    vector<int> G[n];
    rep(i, m){
        int a, b;
        scanf("%d%d", &a, &b);
        --a; --b;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    int d[n][2];
    rep(i, n)rep(j, 2) d[i][j] = mod;
    d[s][0] = 0;
    queue<pii> que;
    que.emplace(s, 0);
    while(!que.empty()){
        int v, p;
        tie(v, p) = que.front();
        que.pop();
        for(int nv: G[v])if(chmin(d[nv][p^1], d[v][p]+1)){
            que.emplace(nv, p^1);
        }
    }

    if(d[t][0] == 2){
        printf("%d\n", 2);
    }
    else if(d[t][1] == mod){
        printf("%d\n", -1);
    }
    else{
        sieve(MAX);
        for(int i=d[t][1]; i<MAX; i+=2){
            if(is_prime(i)){
                printf("%d\n", i);
                return 0;
            }
        }
    }
}