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\nsys.setrecursionlimit(10000000)\ninput = sys.stdin.readline\nfrom math import log\nfrom collections import deque\n\ndef solve():\n    N, Q = list(map(int, input().split()))\n    es = [[] for i in range(N+1)]\n    for i in range(N-1):\n        u, v, c, d = list(map(int, input().split()))\n        es[u].append([v, c, d])\n        es[v].append([u, c, d])\n    LOG_N = int(log(N, 2)) + 1\n    parent = [[-1] * (N+1) for i in range(LOG_N)]\n    level = [None] * (N+1)\n    level[1] = 0\n    dq = deque()\n    dq.append(1)\n    while dq:\n        v = dq.popleft()\n        p = parent[0][v]\n        lv = level[v]\n        for u, c, d in es[v]:\n            if u != p:\n                parent[0][u] = v\n                level[u] = lv + 1\n                dq.append(u)\n    for k in range(LOG_N-1):\n        parentk = parent[k]\n        for v in range(1, N+1):\n            if parentk[v] < 0:\n                parent[k+1][v] = -1\n            else:\n                parent[k+1][v] = parentk[parentk[v]]\n\n    def lca(u, v):\n        if level[u] > level[v]:\n            u, v = v, u\n        for k in range(LOG_N)[::-1]:\n            if (level[v] - level[u]) >> k & 1:\n                v = parent[k][v]\n        if u == v:\n            return u\n        for k in range(LOG_N)[::-1]:\n            if parent[k][u] != parent[k][v]:\n                u, v = parent[k][u], parent[k][v]\n        return parent[0][u]\n\n    qs = [[] for i in range(N+1)]\n    for i in range(Q):\n        x, y, u, v = list(map(int, input().split()))\n        qs[u].append([i, x, y, 1])\n        qs[v].append([i, x, y, 1])\n        a = lca(u, v)\n        qs[a].append([i, x, y, -2])\n    ds = [0] * N\n    cnt = [0] * N\n    query = [0] * Q\n\n    def EulerTour(v, p, d):\n        for id, col, y, f in qs[v]:\n            query[id] += f * (d + y * cnt[col] - ds[col])\n        for u, c, _d in es[v]:\n            if u != p:\n                ds[c] += _d\n                cnt[c] += 1\n                EulerTour(u, v, d + _d)\n                ds[c] -= _d\n                cnt[c] -= 1\n\n    EulerTour(1, -1, 0)\n    for res in query:\n        print(res)\n\ndef __starting_point():\n    solve()\n\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 16453