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

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput = sys.stdin.readline\n\nINF = 10**18\nmod = 10**9+7\n# AOJ GRL_1_A Single Sourse Shortest Path\nimport heapq as hp\n\nN, M = map(int, input().split())\ns, t = map(int, input().split())\ns -= 1; t -= 1\ngraph = [[] for _ in range(N)]\nfor _ in range(M):\n    n, m, d = map(int, input().split())\n    graph[n-1].append((d, m-1)) \n    graph[m-1].append((d, n-1)) #無向グラフならこれ使う\n\ndef dijkstra(s):\n    D = [INF for _ in range(N)] # 頂点iへの最短距離がD[i]\n    B = [0]*N\n    D[s] = 0\n    B[s] = 1\n    q = [] # しまっていく優先度付きキュー\n    hp.heappush(q, (0, s))\n\n    while q:\n        nd, np = hp.heappop(q)\n        if D[np] < nd:\n            continue\n        for d, p in graph[np]:\n            if D[p] > D[np] + d:\n                D[p] = D[np] + d\n                B[p] = B[np]\n                hp.heappush(q, (D[p], p))\n            elif D[p] == D[np] + d:\n                B[p] = (B[p] + B[np])%mod\n    return D, B\n\nsD, sB = dijkstra(s)\ntD, tB = dijkstra(t)\n\ndistance = sD[t]\n\nans = 0\nfor n in range(N):\n    for d, m in graph[n]:\n        if sD[n] + d + tD[m] == distance and 2*sD[n] < distance < 2*(sD[n]+d):\n            ans = (ans + (sB[n]*tB[m])**2)%mod\n    if sD[n] + tD[n] == distance and 2*sD[n] == distance:\n        ans = (ans + (sB[n]*tB[n])**2) % mod\nans = (sB[t]*tB[s] - ans) % mod\nprint(ans)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13825