因为边的数目很少，只有5000，那么我们不妨对边的权值排序，然后5000*5000的进行枚举即可，判断连通性就可以了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
bool operator < (pii a, pii b) { return 1LL * a.first * b.second < 1LL * b.first * a.second; }
pii min(pii a, pii b) { return a < b ? a : b; }
const int maxN = 505;
int N, M, root[maxN], S, T;
int fid(int x) { return x == root[x] ? x : root[x] = fid(root[x]); }
void init() { for(int i = 1; i <= N; i ++) root[i] = i; }
vector<pii> vt[30005];
int lsan[5005];
int main()
{
    scanf("%d%d", &N, &M);
    for(int i = 1, u, v, w; i <= M; i ++)
    {
        scanf("%d%d%d", &u, &v, &w); lsan[i] = w;
        vt[w].push_back(MP(u, v));
    }
    scanf("%d%d", &S, &T);
    sort(lsan + 1, lsan + M + 1);
    int _UP = (int)(unique(lsan + 1, lsan + M + 1) - lsan - 1);
    pii ans = MP(INF, 1);
    for(int i = 1, j, u, v; i <= _UP; i ++)
    {
        init();
        for(j = i; j <= _UP; j ++)
        {
            for(pii edge : vt[lsan[j]])
            {
                u = edge.first;
                v = edge.second;
                u = fid(u);
                v = fid(v);
                if(u ^ v) root[u] = v;
            }
            if(fid(S) == fid(T)) break;
        }
        if(fid(S) != fid(T)) break;
        ans = min(ans, MP(lsan[j] / gcd(lsan[j], lsan[i]), lsan[i] / gcd(lsan[j], lsan[i])));
    }
    if(ans.first == INF) printf("IMPOSSIBLE\n");
    else
    {
        if(ans.second == 1) printf("%d\n", ans.first);
        else printf("%d/%d\n", ans.first, ans.second);
    }
    return 0;
}