Problem: There is a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and Vertex b_i, and the color and length of that edge are c_i and d_i, respectively. Here the color of each edge is represented by an integer between 1 and N-1 (inclusive). The same integer corresponds to the same color, and different integers correspond to different colors.

Answer the following Q queries:

* Query j (1 \leq j \leq Q): assuming that the length of every edge whose color is x_j is changed to y_j, find the distance between Vertex u_j and Vertex v_j. (The changes of the lengths of edges do not affect the subsequent queries.)

Constraints

* 2 \leq N \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq a_i, b_i \leq N
* 1 \leq c_i \leq N-1
* 1 \leq d_i \leq 10^4
* 1 \leq x_j \leq N-1
* 1 \leq y_j \leq 10^4
* 1 \leq u_j < v_j \leq N
* The given graph is a tree.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
a_1 b_1 c_1 d_1
:
a_{N-1} b_{N-1} c_{N-1} d_{N-1}
x_1 y_1 u_1 v_1
:
x_Q y_Q u_Q v_Q


Output

Print Q lines. The j-th line (1 \leq j \leq Q) should contain the answer to Query j.

Example

Input

5 3
1 2 1 10
1 3 2 20
2 4 4 30
5 2 1 40
1 100 1 4
1 100 1 5
3 1000 3 4


Output

130
200
60

Solution (in PYTHON3):

Solution:
def f_colorful_tree(N, Q, Edges, Queries):
    # 参考: https://atcoder.jp/contests/abc133/submissions/6316420
    # https://atcoder.jp/contests/abc133/submissions/6321437

    import sys
    sys.setrecursionlimit(10**5)

    # 入力は1-basedだが、頂点と色の番号については0-basedで考える(以降も同様)
    tree = [[] for _ in range(N)]
    for a, b, c, d in Edges:
        tree[a - 1].append((b - 1, c - 1, d))
        tree[b - 1].append((a - 1, c - 1, d))

    """Least Common Ancestorを求めるための前処理"""
    depth_from_root = [0] * N
    # [k][v]: 頂点vから2**k個根に近づいたときの頂点番号(根の親は-1とする)
    # 頂点数Nが最大 10**5 なので、2**k >= 10**5 となる k=17 までリストを確保する
    parent_to_root = [[-1] * N for _ in range(18)]
    distance_from_root = [0] * N

    def dfs1(parent, current, depth, distance):
        depth_from_root[current] = depth
        parent_to_root[0][current] = parent
        distance_from_root[current] = distance
        for child, _, dist in tree[current]:  # 色についての情報は必要ない
            if child != parent:
                dfs1(current, child, depth + 1, distance + dist)

    dfs1(-1, 0, 0, 0)
    for i in range(17):
        for j in range(N):
            parent_to_root[i + 1][j] = parent_to_root[i][parent_to_root[i][j]]

    def least_common_ancestor(a, b):
        # editorialと同じことをやる
        if depth_from_root[a] > depth_from_root[b]:
            a, b = b, a
        for i in range(18):
            if (depth_from_root[b] - depth_from_root[a]) & (1 << i):
                b = parent_to_root[i][b]
        if a == b:
            return a
        for i in range(18)[::-1]:
            if parent_to_root[i][a] != parent_to_root[i][b]:
                a = parent_to_root[i][a]
                b = parent_to_root[i][b]
        return parent_to_root[0][a]

    """先読みした(オフライン)クエリに対して解を求める"""
    ans = [0] * Q
    v_need = [[] for i in range(N)]
    for i, (x, y, u, v) in enumerate(Queries):
        # 色xの辺の距離を一律yに変更したとき、頂点u-v間の距離を求める
        lca = least_common_ancestor(u - 1, v - 1)
        # 色の情報を必要としているのは u, v, lca の3点
        # それぞれの点に対して以下の情報を格納する
        # (何番目のクエリか, どの色の距離を変えるか, 変える距離はいくらか, 係数)
        # 係数は u-v間の距離 = 根-u + 根-v - 2 * 根-LCA(u, v) であることを反映
        v_need[u - 1].append((i, x - 1, y, 1))
        v_need[v - 1].append((i, x - 1, y, 1))
        v_need[lca].append((i, x - 1, y, -2))

    color_appear = [0] * N  # [c]: 今訪れている頂点までに色cは何回現れたか
    color_distance = [0] * N  # [c]: 今訪れている頂点まででの色cの辺の距離の和

    def dfs2(parent, current):
        nonlocal color_distance
        for i, x, y, coefficient in v_need[current]:
            ans[i] += (coefficient * (distance_from_root[current]
                                      - color_distance[x] + y * color_appear[x]))
        for child, x, y in tree[current]:
            if child == parent:
                continue
            color_appear[x] += 1
            color_distance[x] += y
            dfs2(current, child)
            # 例えば a→b→c→b→d と移動したことを a→b→d とみなす
            # (オイラーツアーのようにする)ため、戻ってきたなら引く
            color_appear[x] -= 1
            color_distance[x] -= y

    dfs2(-1, 0)
    return '\n'.join(map(str, ans))

N, Q = [int(i) for i in input().split()]
Edges = [[int(i) for i in input().split()] for j in range(N - 1)]
Queries = [[int(i) for i in input().split()] for j in range(Q)]
print(f_colorful_tree(N, Q, Edges, Queries))