Problem:
Score : 700 points

Problem Statement

We have a graph with N vertices and M edges, and there are two people on the graph: Takahashi and Aoki.

The i-th edge connects Vertex U_i and Vertex V_i.
The time it takes to traverse this edge is D_i minutes, regardless of direction and who traverses the edge (Takahashi or Aoki).

Takahashi departs Vertex S and Aoki departs Vertex T at the same time. Takahashi travels to Vertex T and Aoki travels to Vertex S, both in the shortest time possible.
Find the number of the pairs of ways for Takahashi and Aoki to choose their shortest paths such that they never meet (at a vertex or on an edge) during the travel, modulo 10^9 + 7.

Constraints

1 \leq N \leq 100 000

1 \leq M \leq 200 000

1 \leq S, T \leq N

S \neq T

1 \leq U_i, V_i \leq N (1 \leq i \leq M)

1 \leq D_i \leq 10^9 (1 \leq i \leq M)

If i \neq j, then (U_i, V_i) \neq (U_j, V_j) and (U_i, V_i) \neq (V_j, U_j).

U_i \neq V_i (1 \leq i \leq M)

D_i are integers.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
S T
U_1 V_1 D_1
U_2 V_2 D_2
:
U_M V_M D_M

Output

Print the answer.

Sample Input 1

4 4
1 3
1 2 1
2 3 1
3 4 1
4 1 1

Sample Output 1

2

There are two ways to choose shortest paths that satisfies the condition:

Takahashi chooses the path 1 \rightarrow 2 \rightarrow 3, and Aoki chooses the path 3 \rightarrow 4 \rightarrow 1.

Takahashi chooses the path 1 \rightarrow 4 \rightarrow 3, and Aoki chooses the path 3 \rightarrow 2 \rightarrow 1.

Sample Input 2

3 3
1 3
1 2 1
2 3 1
3 1 2

Sample Output 2

2

Sample Input 3

3 3
1 3
1 2 1
2 3 1
3 1 2

Sample Output 3

2

Sample Input 4

8 13
4 2
7 3 9
6 2 3
1 6 4
7 6 9
3 8 9
1 2 2
2 8 12
8 6 9
2 5 5
4 2 18
5 3 7
5 1 515371567
4 8 6

Sample Output 4

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
using ll = long long;
using u64 = uint_fast64_t;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007;
//constexpr long long MOD = 998244353;
template<class T1, class T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return true;} return false; }
template<class T1, class T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return true;} return false; }
inline int popcount(int x) {return __builtin_popcount(x);}
inline int popcount(long long x) {return __builtin_popcountll(x);}
void print() { cout << "\n"; }
template<class T, class... Args>
void print(const T &x, const Args &... args) {
    cout << x << " ";
    print(args...);
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////////

template<typename T>
struct edge {
    int src,to;
    T cost;

    edge(){};
    edge(int to, T cost) : src(-1), to(to), cost(cost) {}
    edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}
    bool operator<(const edge &e) const {
        return cost < e.cost;
    }
};

template <uint_fast64_t Modulus> 
struct ModInt {
    using u64 = uint_fast64_t;

    u64 a;

    constexpr ModInt(const long long x = 0) noexcept : a(x >= 0 ? x % Modulus : (Modulus - (-x) % Modulus) % Modulus) {}
    constexpr u64 &value() noexcept {return a;}
    constexpr const u64 &value() const noexcept {return a;}
    constexpr ModInt operator+(const ModInt rhs) const noexcept {return ModInt(*this) += rhs;}
    constexpr ModInt operator-(const ModInt rhs) const noexcept {return ModInt(*this) -= rhs;}
    constexpr ModInt operator*(const ModInt rhs) const noexcept {return ModInt(*this) *= rhs;}
    constexpr ModInt operator/(const ModInt rhs) const noexcept {return ModInt(*this) /= rhs;}
    constexpr ModInt operator^(const long long rhs) const noexcept {return ModInt(*this) ^= rhs;}
    constexpr bool operator==(const ModInt &rhs) const noexcept {return a == rhs.a;}
    constexpr bool operator!=(const ModInt &rhs) const noexcept {return a != rhs.a;}
    constexpr ModInt &operator+=(const ModInt rhs) noexcept {
        a += rhs.a;
        if (a >= Modulus) {
            a -= Modulus;
        }
        return *this;
    }
    constexpr ModInt &operator-=(const ModInt rhs) noexcept {
        if (a < rhs.a) {
            a += Modulus;
        }
        a -= rhs.a;
        return *this;
    }
    constexpr ModInt &operator*=(const ModInt rhs) noexcept {
        a = a * rhs.a % Modulus;
        return *this;
    }
    constexpr ModInt &operator/=(ModInt rhs) noexcept {
        u64 exp = Modulus - 2;
        while (exp) {
            if (exp&1) *this *= rhs;
            exp >>= 1;
            rhs *= rhs;
        }
        return *this;
    }
    constexpr ModInt &operator^=(long long exp) noexcept {
        ModInt rhs = a;
        a = 1;
        while (exp) {
            if (exp&1) *this *= rhs;
            exp >>= 1;
            rhs *= rhs;
        }
        return *this;
    }

    friend ostream &operator<<(ostream& os, const ModInt& rhs) noexcept {return os << rhs.a;}
    friend istream &operator>>(istream& is, ModInt& rhs) noexcept {long long a; is >> a; rhs = a; return is;}
};

using mint = ModInt<MOD>;

template<typename T>
pair<vector<T>,vector<mint>> Dijkstra(vector<vector<edge<T>>> &G, int s, T inf) {
    vector<T> dist(G.size(),inf);
    vector<mint> cnt(G.size());
    using Pi = pair<T, int>;
    priority_queue<Pi, vector<Pi>, greater<Pi>> pq;
    dist[s] = 0;
    cnt[s] = 1;
    pq.emplace(dist[s],s);
    while (!pq.empty()) {
        T d; int v;
        tie(d,v) = pq.top(); pq.pop();
        if (d > dist[v]) continue;
        for (auto &e : G[v]) {
            if (dist[e.to] > dist[v] + e.cost) {
                dist[e.to] = dist[v] + e.cost;
                pq.emplace(dist[e.to],e.to);
                cnt[e.to] = cnt[v];
            } else if (dist[e.to] == dist[v] + e.cost) {
                cnt[e.to] += cnt[v];
            }
        }
    }

    return make_pair(dist,cnt);
}

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int N,M; cin >> N >> M;
    int s,t; cin >> s >> t;
    --s; --t;
    vector<vector<edge<ll>>> G(N);
    vector<edge<ll>> E(M);
    rep(i,M) {
        int u,v,w; cin >> u >> v >> w;
        u--; v--;
        G[u].emplace_back(v,w);
        G[v].emplace_back(u,w);
        E[i] = edge<ll>(u,v,w);
    }

    vector<ll> dist1,dist2;
    vector<mint> cnt1,cnt2;
    tie(dist1,cnt1) = Dijkstra<ll>(G,s,1e18);
    tie(dist2,cnt2) = Dijkstra<ll>(G,t,1e18);
    mint ans = cnt1[t]*cnt2[s];
    rep(i,M) {
        int u = E[i].src;
        int v = E[i].to;
        if (dist1[u] > dist1[v]) swap(u,v);
        if (dist1[u]+E[i].cost != dist1[v]) continue;
        if (dist1[u]+dist2[v]+E[i].cost==dist1[t] and dist1[u] < dist2[u] and dist1[v] > dist2[v]) {
            ans -= cnt1[u]*cnt1[u]*cnt2[v]*cnt2[v];
        }
    }

    rep(i,N) {
        if (dist1[i]+dist2[i]==dist1[t] and dist1[i]==dist2[i]) {
            ans -= cnt1[i]*cnt1[i]*cnt2[i]*cnt2[i];
        }
    }

    cout << ans << ln;
}