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 heapq
import sys
input = sys.stdin.buffer.readline

def dijkstra(start: int, graph: list) -> list:
    """dijkstra法: 始点startから各頂点への最短距離を求める
    計算量: O((E+V)logV)
    """
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    n = len(graph)
    dist = [INF] * n
    ptn = [0] * n 
    dist[start] = 0
    ptn[start] = 1 
    q = [(0, start)] # q = [(startからの距離, 現在地)]
    while q:
        d, v = heapq.heappop(q)
        if dist[v] < d:
            continue
        for nxt_v, cost in graph[v]:
            if dist[v] + cost == dist[nxt_v]:
                ptn[nxt_v] += ptn[v]
                ptn[nxt_v] %= MOD
            if dist[v] + cost < dist[nxt_v]:
                dist[nxt_v] = dist[v] + cost
                ptn[nxt_v] = ptn[v]
                ptn[nxt_v] %= MOD
                heapq.heappush(q, (dist[nxt_v], nxt_v))
    return dist, ptn




n, m = list(map(int, input().split()))
s, t = list(map(int, input().split()))
info = [list(map(int, input().split())) for i in range(m)]
MOD = 10 ** 9 + 7

s -= 1
t -= 1
graph = [[] for i in range(n)]
for a, b, cost in info:
    a -= 1
    b -= 1
    graph[a].append((b, cost))
    graph[b].append((a, cost))
    
dist_st, ptn_st = dijkstra(s, graph)
dist_ts, ptn_ts = dijkstra(t, graph)

min_cost, max_ptn = dist_st[t], ptn_st[t]
all_ptn = (max_ptn * max_ptn) % MOD

# 最短路で使用する頂点を復元
use = [False] * n
use[t] = True
stack = [t]
while stack:
    v = stack.pop()
    for prv_v, cost in graph[v]:
        if use[prv_v]:
            continue
        if dist_st[v] == dist_st[prv_v] + cost:
            use[prv_v] = True
            stack.append(prv_v)

# 頂点上で出会うとき
for i in range(n):
    if dist_st[i] == dist_ts[i] and use[i]:
        all_ptn -= (ptn_st[i] * ptn_ts[i]) ** 2
        all_ptn %= MOD
        
# 辺上で出会うとき
for a, b, cost in info:
    a -= 1
    b -= 1
    if abs(dist_st[a] - dist_st[b]) != cost:
        continue
    if (not use[a]) or (not use[b]):
        continue
    if dist_st[a] > dist_st[b]:    
        a, b = b, a
    if dist_st[a] < dist_ts[a] and dist_st[b] > dist_ts[b]:
        all_ptn -= (ptn_st[a] * ptn_ts[b]) ** 2
    all_ptn %= MOD
print(all_ptn)