Problem: 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.

Examples

Input

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


Output

2


Input

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


Output

2


Input

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


Output

6

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
from heapq import *

"""
とりあえず何通りあるかを数える：各頂点を通る時点での場合の数
最後に衝突する方法を除外する：頂点での衝突、辺での衝突
辺での衝突：距離を見れば分かる
"""

MOD = 10 ** 9 + 7

N,M = map(int,input().split())
S,T = map(int,input().split())
graph = [[] for _ in range(N+1)]
for _ in range(M):
    u,v,d = map(int,input().split())
    graph[u].append((v,d))
    graph[v].append((u,d))

def dijkstra(S):
    # 距離、パスの個数を返す
    INF = 10 ** 15
    dist = [INF] * (N+1)
    cnt_path = [0] * (N+1)
    cnt_path[S] = 1
    q = [(0,S)] # 距離、場所
    while q:
        d,x = heappop(q)
        if dist[x] < d:
            continue
        dist[x] = d
        for y,dy in graph[x]:
            if d + dy > dist[y]:
                continue
            if d + dy == dist[y]:
                cnt_path[y] += cnt_path[x]
                continue
            cnt_path[y] = cnt_path[x]
            dist[y] = d + dy
            heappush(q,(d+dy,y))
    return dist, cnt_path

dist_S, cnt_path_S = dijkstra(S)
dist_T, cnt_path_T = dijkstra(T)

D = dist_S[T]
# 最短路の組
answer = cnt_path_S[T] * cnt_path_T[S]
# 頂点での衝突
for x in range(1,N+1):
    if dist_S[x] == dist_T[x]:
        t = cnt_path_S[x] * cnt_path_T[x] % MOD
        answer -= t * t % MOD
# 辺での衝突。S側がx、T側がy
for x in range(1,N+1):
    for y,d in graph[x]:
        dx = dist_S[x]
        dy = dist_T[y]
        if dx + dy + d == D and dx + d > dy and dy + d > dx:
            t = cnt_path_S[x] * cnt_path_T[y] % MOD
            answer -= t * t % MOD

answer %= MOD
print(answer)