#ifndef DEBUG
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("tune=native")
#endif

#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

using ll = long long;
using mint = modint998244353;
// using mint = modint1000000007;

template <typename T> using vec = vector<T>;
template <typename T> using pr = pair<T, T>;
template <typename T> using mipq = priority_queue<T, vec<T>, greater<T>>;

#define overload4(_1, _2, _3, _4, name, ...) name
#define rep1(i, n) for (auto i = decay_t<decltype(n)>{}; i < (n); i++)
#define rep2(i, l, r) for (auto i = (l); i < (r); i++)
#define rep3(i, l, r, d) for (auto i = (l); i < (r); i += (d))
#define rep(...) overload4(__VA_ARGS__, rep3, rep2, rep1)(__VA_ARGS__)
#define rrep(i, r, l) for (auto i = (r); i >= (l); i--)
template <class T> bool chmax(T& a, const T& b) { return a < b ? a = b, true : false; }
template <class T> bool chmin(T& a, const T& b) { return a > b ? a = b, true : false; }

constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;

void solve();

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cout << fixed << setprecision(15);
    int T = 1;
    // cin >> T;
    while (T--) solve();
}

// 行列累乗しろってことー？

using mat = vec<vec<mint>>;

mat mul(const mat& a, const mat& b) {
    int n = a.size();
    mat res(n, vec<mint>(n));
    rep(i, n) rep(k, n) rep(j, n) res[i][j] += a[i][k] * b[k][j];
    return res;
}

mat pow(mat x, ll k) {
    int n = x.size();
    mat res(n, vec<mint>(n));
    rep(i, n) res[i][i] = 1;
    while (k) {
        if (k & 1) res = mul(res, x);
        x = mul(x, x);
        k >>= 1;
    }
    return res;
}

void solve() {
    int N, M;
    cin >> N >> M;
    vec<vec<int>> G(N);
    rep(i, M) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }

    mat trans(N, vec<mint>(N));
    rep(u, N) {
        mint p = mint(G[u].size()).inv();
        for (auto v : G[u]) trans[u][v] = p;
    }

    ll S, T; int A, B;
    cin >> S >> T >> A >> B;
    S--; T--; A--; B--;
    mint res = pow(trans, T)[0][B] * pow(trans, S - T)[B][A] / pow(trans, S)[0][A];
    cout << res.val() << '\n';
}