Mr. Kitayuta has just bought an undirected graph with n vertices and m edges. The vertices of the graph are numbered from 1 to n. Each edge, namely edge i, has a color ci, connecting vertex ai and bi.

Mr. Kitayuta wants you to process the following q queries.

In the i-th query, he gives you two integers - ui and vi.

Find the number of the colors that satisfy the following condition: the edges of that color connect vertex ui and vertex vi directly or indirectly.

Input

The first line of the input contains space-separated two integers - n and m(2 ≤ n ≤ 105, 1 ≤ m ≤ 105), denoting the number of the vertices and the number of the edges, respectively.

The next m lines contain space-separated three integers - ai, bi(1 ≤ ai < bi ≤ n) and ci(1 ≤ ci ≤ m). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if i ≠ j, (ai, bi, ci) ≠ (aj, bj, cj).

The next line contains a integer- q(1 ≤ q ≤ 105), denoting the number of the queries.

Then follows q lines, containing space-separated two integers - ui and vi(1 ≤ ui, vi ≤ n). It is guaranteed that ui ≠ vi.

Output

For each query, print the answer in a separate line.

Examples

Input

4 5
1 2 1
1 2 2
2 3 1
2 3 3
2 4 3
3
1 2
3 4
1 4


Output

2
1
0


Input

5 7
1 5 1
2 5 1
3 5 1
4 5 1
1 2 2
2 3 2
3 4 2
5
1 5
5 1
2 5
1 5
1 4


Output

1
1
1
1
2

Note

Let's consider the first sample. 

<image> The figure above shows the first sample. 

  * Vertex 1 and vertex 2 are connected by color 1 and 2. 
  * Vertex 3 and vertex 4 are connected by color 3. 
  * Vertex 1 and vertex 4 are not connected by any single color. 
def gerarGrafo(cor):
    adjancencias = {}
    for aresta in arestas:
        if aresta[2] == cor:
            if adjancencias.get(aresta[0]) == None:
                adjancencias[aresta[0]] = []
            if adjancencias.get(aresta[1]) == None:
                adjancencias[aresta[1]] = []
            adjancencias[aresta[0]].append(aresta[1])
            adjancencias[aresta[1]].append(aresta[0])
    return adjancencias

def dfs(u, w, adjacencias):
    pilha = [u]
    visitados = {u: 1}
    existeCaminhoUV = False
    if adjacencias.get(u) == None or adjacencias.get(w) == None:
        return existeCaminhoUV
    
    while len(pilha) > 0:
        vertice = pilha.pop()
        for v in adjacencias[vertice]:
            if visitados.get(v) == None:
                visitados[v] = 1
                pilha.append(v)
    if visitados.get(w) != None:
        existeCaminhoUV = True
    return existeCaminhoUV
    

entrada = input().split()
n = int(entrada[0])
m = int(entrada[1])
cores = set()
arestas = []

i = 0
while i < m:
    aresta = input().split()
    cores.add(aresta[2])
    arestas.append(aresta)
    i += 1

q = int(input())
paresVertices = []

i = 0
while i < q:
    paresVertices.append(input().split())
    i += 1

res = [0] * q
for cor in cores:
    grafo = gerarGrafo(cor)
    i = 0
    for par in paresVertices:
        if dfs(par[0], par[1], grafo):
            res[i] += 1
        i += 1

for e in res:
    print(e)