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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import heapq
import sys
def input():
	return sys.stdin.readline()[:-1]

MOD = 10**9+7
class DijkstraList():
	#隣接リスト版
	#同一頂点の複数回探索を防ぐため訪問した頂点数を変数cntで持つ
	def __init__(self, adj, start):
		self.list = adj
		self.start = start
		self.size = len(adj)

	def solve(self):
		self.dist = [float("inf") for _ in range(self.size)]
		self.dist[self.start] = 0
		self.dp = [0 for _ in range(self.size)]
		self.dp[self.start] = 1
		#self.prev = [-1 for _ in range(self.size)]
		self.q = []
		self.cnt = 0

		heapq.heappush(self.q, (0, self.start))

		while self.q and self.cnt < self.size:
			u_dist, u = heapq.heappop(self.q)
			if self.dist[u] < u_dist:
				continue
			for v, w in self.list[u]:
				if self.dist[v] > u_dist + w:
					self.dist[v] = u_dist + w
					self.dp[v] = self.dp[u]
					#self.prev[v] = u
					heapq.heappush(self.q, (self.dist[v], v))
				elif self.dist[v] == u_dist + w:
					self.dp[v] += self.dp[u]
					self.dp[v] %= MOD

			self.cnt += 1
		#print(self.dp)
		return

	def distance(self, goal):
		return self.dist[goal]

	def get_dp(self, x):
		return self.dp[x]

n, m = map(int, input().split())
s, t = map(int, input().split())
s, t = s-1, t-1
adj = [[] for _ in range(n)]
for _ in range(m):
	u, v, d = map(int, input().split())
	adj[u-1].append([v-1, d])
	adj[v-1].append([u-1, d])
S, T = DijkstraList(adj, s), DijkstraList(adj, t)
S.solve()
T.solve()
ans = pow(S.get_dp(t), 2, MOD)
goal = S.distance(t)
#print(ans, goal)

for i in range(n):
	SI, TI = S.distance(i), T.distance(i)
	if SI + TI != goal:
		continue
	if SI*2 == goal:
		ans -= S.get_dp(i) * T.get_dp(i) * S.get_dp(i) * T.get_dp(i)
		ans %= MOD

	else:
		for j, d in adj[i]:
			if i>j:
				continue
			SJ, TJ = S.distance(j), T.distance(j)
			if SJ + TJ != goal or abs(SI - SJ) != d:
				continue
			if SI*2 < goal < SJ*2:
				#print(i, j, SI, SJ)
				ans -= S.get_dp(i) * T.get_dp(j) * S.get_dp(i) * T.get_dp(j)
				ans %= MOD
			elif SJ*2 < goal < SI*2:
				#print(i, j, SI, SJ)
				ans -= S.get_dp(j) * T.get_dp(i) * S.get_dp(j) * T.get_dp(i)
				ans %= MOD

print(ans)