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 = 'from collections import defaultdict\nfrom math import log2\nimport sys\nsys.setrecursionlimit(10**6)\ninput=sys.stdin.readline\n\ndef f(n,q):\n    def dfs(i, oi=-1, dpt=0):\n        depth[i] = dpt\n        parent_2k[0][i] = oi\n        for ki, c, d in to[i]:\n            if ki == oi: continue\n            dfs(ki, i, dpt + 1)\n        return\n\n    def make_parent(level):\n        parent_2kk1 = parent_2k[0]\n        for k in range(1, level):\n            parent_2kk = parent_2k[k]\n            for i in range(n):\n                parent1 = parent_2kk1[i]\n                if parent1 != -1:\n                    parent_2kk[i] = parent_2kk1[parent1]\n            parent_2kk1 = parent_2kk\n        return\n\n    def lca(u, v):\n        dist_uv = depth[u] - depth[v]\n        if dist_uv < 0:\n            u, v = v, u\n            dist_uv *= -1\n\n        k = 0\n        while dist_uv:\n            if dist_uv & 1:\n                u = parent_2k[k][u]\n            dist_uv >>= 1\n            k += 1\n\n        if u == v:\n            return u\n\n        for k in range(int(log2(depth[u])) + 1, -1, -1):\n            pu = parent_2k[k][u]\n            pv = parent_2k[k][v]\n            if pu != pv:\n                u = pu\n                v = pv\n        return parent_2k[0][u]\n\n    to = defaultdict(list)\n    root = 0\n    depth = [0] * n\n    level = int(log2(n)) + 1\n    parent_2k = [[-1] * n for _ in range(level)]\n\n    for _ in range(n - 1):\n        a, b, c, d = list(map(int, input().split()))\n        to[a - 1] += [[b - 1, c - 1, d]]\n        to[b - 1] += [[a - 1, c - 1, d]]\n\n    dfs(root)\n    make_parent(level)\n\n    qs = [[] for _ in range(n)]\n    for qi in range(q):\n        x, y, u, v = list(map(int, input().split()))\n        qs[u - 1].append([qi, x - 1, y, 1])\n        qs[v - 1].append([qi, x - 1, y, 1])\n        qs[lca(u - 1, v - 1)].append([qi, x - 1, y, -2])\n\n    cnt_c = [0] * n\n    sum_c = [0] * n\n    ans = [0] * q\n\n    def dfs2(i, oi=-1, sum_d=0):\n        for qi, qc, qd, qk in qs[i]:\n            ans[qi] += (sum_d - sum_c[qc] + qd * cnt_c[qc]) * qk\n        for ki, c, d in to[i]:\n            if ki == oi: continue\n            cnt_c[c] += 1\n            sum_c[c] += d\n            dfs2(ki, i, sum_d + d)\n            cnt_c[c] -= 1\n            sum_c[c] -= d\n        return\n\n    dfs2(root)\n    for x in ans:\n        print(x)\n\nn, q = list(map(int, input().split()))\nf(n,q)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13858