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 = sys.stdin.readline\nsys.setrecursionlimit(10**7)\n\nclass LCA:\n    def __init__(self, edge, root):\n        self.l = len(edge)\n        self.stack = []\n        self.stack.append((root, 0))\n        self.vs = []\n        self.depth = [0]*self.l\n        self.firstVisited = [-1]*self.l\n        while self.stack:\n            v, d = self.stack[-1]\n            if self.firstVisited[v] == -1:\n                self.firstVisited[v] = len(self.vs)\n            self.vs.append(v)\n            self.depth[v] = d\n            if edge[v]:\n                self.stack.append((edge[v].pop(), d+1))\n            else:\n                self.stack.pop()\n        \n        self.n = len(self.vs)\n        self.INF = (self.l, self.l)\n        self.N0 = 1 << (self.n-1).bit_length()\n        self.seg = [self.INF]*(self.N0 << 1)\n        for i, v in enumerate(self.vs, self.N0):\n            self.seg[i] = (self.depth[v], v)\n        for i in range(self.N0-1, 0, -1):\n            self.seg[i] = min(self.seg[2*i], self.seg[2*i+1])\n    \n    def _query(self, l, r):\n        res = self.INF\n        l += self.N0\n        r += self.N0\n        while l < r:\n            if r & 1:\n                res = min(res, self.seg[r-1])\n            if l & 1:\n                res = min(res, self.seg[l])\n                l += 1\n            l >>= 1\n            r >>= 1\n     \n        return res\n\n    def query(self, u, v):\n        fu, fv = self.firstVisited[u], self.firstVisited[v]\n        if fu > fv:\n            fu, fv = fv, fu\n        return self._query(fu, fv+1)[1]\n\nn, q = map(int, input().split())\nT = [[] for _ in range(n)]\nfor _ in range(n-1):\n  a, b, c, d = map(int, input().split())\n  a -= 1\n  b -= 1\n  T[a].append((b, c, d))\n  T[b].append((a, c, d))\nG = [[] for _ in range(n)]\nedge = [[] for _ in range(n)]\nstack = [(0, -1)]\nwhile stack:\n  v, par = stack.pop()\n  for nv, c, d in T[v]:\n    if nv == par:\n      continue\n    G[v].append((nv, c, d))\n    edge[v].append(nv)\n    stack.append((nv, v))\n\nseg = LCA(edge, 0)\nA = [0]*q\nC = [[] for _ in range(n)]\nnum = [0]*n\ncum = [0]*n\nfor i in range(q):\n  x, y, u, v = map(int, input().split())\n  u -= 1\n  v -= 1\n  a = seg.query(u, v)\n  C[u].append((i, x, y, 1))\n  C[v].append((i, x, y, 1))\n  C[a].append((i, x, y, -2))\ndist = 0\nstack = [0]\ndef dfs(v):\n  nonlocal dist\n  for i, x, y, b in C[v]:\n    A[i] += (dist + num[x]*y - cum[x]) * b\n  for nv, c, d in G[v]:\n    dist += d\n    num[c] += 1\n    cum[c] += d\n    dfs(nv)\n    dist -= d\n    num[c] -= 1\n    cum[c] -= d\ndfs(0)\nprint(*A, sep="\\n")'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: