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 Ruby (2.3.3) (judge verdict: Wrong Answer)

class Heap
    attr_reader :size
    def initialize
        @heap = []
        @size = 0
    end

    def length
        @size
    end

    def push(x)
        i = @size
        @size += 1
        while i > 0 do
            p = (i - 1) / 2
            break if @heap[p] <= x
            @heap[i] = @heap[p]
            i = p
        end
        @heap[i] = x
        self
    end

    def pop
        return nil if @size == 0
        ret = @heap[0]
        @size -= 1
        x = @heap[@size]
        i = 0
        while i * 2 + 1 < @size do
            a = i * 2 + 1
            b = a + 1
            if b < @size && @heap[b] < @heap[a]
                a = b
            end
            break if @heap[a] >= x
            @heap[i] = @heap[a]
            i = a
        end
        @heap[i] = x
        @heap.pop
        return ret
    end
end

class Dpair
    attr_reader :dist, :vertex
    include Comparable
    def initialize(dist, vertex)
        @dist = dist
        @vertex = vertex
    end
    def <=>(other)
        @dist <=> other.dist
    end
end

# edges[src][dst] = cost
def dijkstra(start, n, edges)
    queue = Heap.new
    dist = Array.new(n+1, Float::INFINITY)
    prev = Array.new(n+1)
    dist[start] = 0
    queue.push(Dpair.new(0, start))
    while queue.size > 0 do
        dpair = queue.pop
        v = dpair.vertex
        next if dist[v] < dpair.dist
        edges[v].each do |dst, cost|
            if dist[dst] > dist[v] + cost
                dist[dst] = dist[v] + cost
                queue.push(Dpair.new(dist[dst], dst))
                prev[dst] = v
            end
        end
    end
    return dist, prev
end

def prev2path(start, goal, prev)
    path = [goal]
    pt = goal
    loop do
        break if pt == start
        pt = prev[pt]
        path.unshift(pt)
    end
    return path
end

N, M = gets.split.map(&:to_i)
S, T = gets.split.map(&:to_i)
edges = Array.new(N){{}}
M.times do
    u, v, d = gets.split.map(&:to_i)
    edges[u-1][v-1] = d
    edges[v-1][u-1] = d
end

dist, prev = dijkstra(S-1, N, edges)