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 = 'import sys\ninput=lambda :sys.stdin.readline().rstrip()\nsys.setrecursionlimit(10**8)\n\nclass LCA(object):\n    def __init__(self, to, root=0):\n        self.__n = len(to)\n        self.__to = to\n        self.__logn = (self.__n - 1).bit_length()\n        self.__depth = [-1] * self.__n\n        self.__dist = [-1] * self.__n\n        self.__depth[root] = 0\n        self.__dist[root] = 0\n        self.__parents = [[-1] * self.__n for _ in range(self.__logn)]\n        self.__dfs(root)\n        self.__doubling()\n\n    def __dfs(self, v):\n        for u, c, d in self.__to[v]:\n            if self.__depth[u] != -1: continue\n            self.__parents[0][u] = v\n            self.__depth[u] = self.__depth[v] + 1\n            self.__dist[u] = self.__dist[v] + d\n            self.__dfs(u)\n\n    def __doubling(self):\n        for i in range(1, self.__logn):\n            for v in range(self.__n):\n                if self.__parents[i - 1][v] == -1: continue\n                self.__parents[i][v] = self.__parents[i - 1][self.__parents[i - 1][v]]\n\n    @property\n    def depth(self):\n        return self.__depth\n\n    @property\n    def dist(self):\n        return self.__dist\n\n    def get(self, u, v):\n        dd = self.__depth[v] - self.__depth[u]\n        if dd < 0:\n            u, v = v, u\n            dd *= -1\n        for i in range(self.__logn):\n            if dd & (2 ** i):\n                v = self.__parents[i][v]\n        if v == u: return v\n        for i in range(self.__logn - 1, -1, -1):\n            pu = self.__parents[i][u]\n            pv = self.__parents[i][v]\n            if pu != pv: u, v = pu, pv\n        return self.__parents[0][u]\n\ndef resolve():\n    n, q = map(int, input().split())\n    to = [[] for _ in range(n)]\n    for _ in range(n-1):\n        a, b, c, d = map(int, input().split())\n        a -= 1; b -= 1\n        to[a].append((b, c, d))\n        to[b].append((a, c, d))\n\n    G = LCA(to)\n    Query = [[] for _ in range(n)]\n    for i in range(q):\n    # idx, color, mag, coef\n        x, y, u, v = map(int, input().split())\n        u -= 1; v -= 1\n        c = G.get(u, v)\n        Query[u].append((i, x, y, 1))\n        Query[v].append((i, x, y, 1))\n        Query[c].append((i, x, y, -2))\n\n    ans = [0] * q\n    S = [[0, 0] for _ in range(n)]\n\n    def dfs(v, p=-1):\n        for idx, color, mag, coef in Query[v]:\n            x = G.dist[v]\n            x -= S[color][1]\n            x += mag * S[color][0]\n            ans[idx] += x * coef\n        for nv, color, d in to[v]:\n            if nv == p: continue\n            S[color][0] += 1\n            S[color][1] += d\n            dfs(nv, v)\n            S[color][0] -= 1\n            S[color][1] -= d\n    dfs(0)\n    print(*ans, sep="\\n")\n\nresolve()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 19047