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.

-----Sample 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

-----Sample Output-----
130
200
60

The graph in this input is as follows:

Here the edges of Color 1 are shown as solid red lines, the edge of Color 2 is shown as a bold green line, and the edge of Color 4 is shown as a blue dashed line.
 - Query 1: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 4 is 100 + 30 = 130.
 - Query 2: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 5 is 100 + 100 = 200.
 - Query 3: Assuming that the length of every edge whose color is 3 is changed to 1000 (there is no such edge), the distance between Vertex 3 and Vertex 4 is 20 + 10 + 30 = 60. Note that the edges of Color 1 now have their original lengths.

```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 = 'def f_colorful_tree(N, Q, Edges, Queries):\n    # 参考: https://atcoder.jp/contests/abc133/submissions/6316420\n    # https://atcoder.jp/contests/abc133/submissions/6321437\n\n    import sys\n    sys.setrecursionlimit(10**5)\n\n    # 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)\n    tree = [[] for _ in range(N)]\n    for a, b, c, d in Edges:\n        tree[a - 1].append((b - 1, c - 1, d))\n        tree[b - 1].append((a - 1, c - 1, d))\n\n    """Least Common Ancestorを求めるための前処理"""\n    depth_from_root = [0] * N\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)\n    # 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\n    parent_to_root = [[-1] * N for _ in range(18)]\n    distance_from_root = [0] * N\n\n    def dfs1(parent, current, depth, distance):\n        depth_from_root[current] = depth\n        parent_to_root[0][current] = parent\n        distance_from_root[current] = distance\n        for child, _, dist in tree[current]:  # 色についての情報は必要ない\n            if child != parent:\n                dfs1(current, child, depth + 1, distance + dist)\n\n    dfs1(-1, 0, 0, 0)\n    for i in range(17):\n        for j in range(N):\n            parent_to_root[i + 1][j] = parent_to_root[i][parent_to_root[i][j]]\n\n    def least_common_ancestor(a, b):\n        # editorialと同じことをやる\n        if depth_from_root[a] > depth_from_root[b]:\n            a, b = b, a\n        for i in range(18):\n            if (depth_from_root[b] - depth_from_root[a]) & (1 << i):\n                b = parent_to_root[i][b]\n        if a == b:\n            return a\n        for i in range(18)[::-1]:\n            if parent_to_root[i][a] != parent_to_root[i][b]:\n                a = parent_to_root[i][a]\n                b = parent_to_root[i][b]\n        return parent_to_root[0][a]\n\n    """先読みした(オフライン)クエリに対して解を求める"""\n    ans = [0] * Q\n    v_need = [[] for i in range(N)]\n    for i, (x, y, u, v) in enumerate(Queries):\n        # 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\n        lca = least_common_ancestor(u - 1, v - 1)\n        # 色の情報を必要としているのは u, v, lca の3点\n        # それぞれの点に対して以下の情報を格納する\n        # (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)\n        # 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\n        v_need[u - 1].append((i, x - 1, y, 1))\n        v_need[v - 1].append((i, x - 1, y, 1))\n        v_need[lca].append((i, x - 1, y, -2))\n\n    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\n    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\n\n    def dfs2(parent, current):\n        nonlocal color_distance\n        for i, x, y, coefficient in v_need[current]:\n            ans[i] += (coefficient * (distance_from_root[current]\n                                      - color_distance[x] + y * color_appear[x]))\n        for child, x, y in tree[current]:\n            if child == parent:\n                continue\n            color_appear[x] += 1\n            color_distance[x] += y\n            dfs2(current, child)\n            # 例えば a→b→c→b→d と移動したことを a→b→d とみなす\n            # (オイラーツアーのようにする)ため、戻ってきたなら引く\n            color_appear[x] -= 1\n            color_distance[x] -= y\n\n    dfs2(-1, 0)\n    return \'\\n\'.join(map(str, ans))\n\nN, Q = [int(i) for i in input().split()]\nEdges = [[int(i) for i in input().split()] for j in range(N - 1)]\nQueries = [[int(i) for i in input().split()] for j in range(Q)]\nprint(f_colorful_tree(N, Q, Edges, Queries))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15432