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**5)\n\nN, Q = map(int, input().split())\n\npath = [[] for _ in range(N)]\n\nfor _ in range(N-1) :\n    a, b, c, d = (int(i) for i in input().split())\n    path[a-1].append((b-1, c-1, d))\n    path[b-1].append((a-1, c-1, d))\n\n# doublingに必要なKを求める\nfor K in range(18) :\n    if 2 ** K >= N :\n        break\n\n# dfs\nparent = [[-1] * N for _ in range(K)]\nrank = [-1 for _ in range(N)]\n\nrank[0] = 0\nqueue = [0]\n\nwhile queue :\n    cur = queue.pop()\n    for nex, _, _ in path[cur] :\n        if rank[nex] < 0 :\n            queue.append(nex)\n            parent[0][nex] = cur\n            rank[nex] = rank[cur] + 1\n\n# doubling        \nfor i in range(1, K) :\n    for j in range(N) :\n        if parent[i-1][j] > 0 :\n            parent[i][j] = parent[i-1][parent[i-1][j]]\n\n# lca\ndef lca(a, b) :\n    if rank[a] > rank[b] :\n        a, b = b, a\n\n    diff = rank[b] - rank[a]\n    i = 0\n    while diff > 0 :\n        if diff & 1 :\n            b = parent[i][b]\n        diff >>= 1\n        i += 1\n        \n    if a == b :\n        return a\n\n    for i in range(K-1, -1, -1) :\n        if parent[i][a] != parent[i][b] :\n            a = parent[i][a]\n            b = parent[i][b]\n\n    return parent[0][a]\n\n# Queryの先読み\nschedule = [[] for _ in range(N)]\nfor i in range(Q) : \n    x, y, u, v = map(int, input().split())\n    x, u, v = x-1, u-1, v-1\n    l = lca(u, v)\n    schedule[u].append((i, 1, x, y))\n    schedule[v].append((i, 1, x, y))\n    schedule[l].append((i, -2, x, y))\n\nret = [0] * Q\nC = [0] * (N-1)\nD = [0] * (N-1)\n\ndef dfs(cur, pre, tot) :\n    for i, t, c, d in schedule[cur] :\n        ret[i] += t * (tot - D[c] + C[c] * d)\n    \n    for nex, c, d in path[cur] :\n        if nex == pre :\n            continue\n        C[c] += 1\n        D[c] += d\n        dfs(nex, cur, tot + d)\n        C[c] -= 1\n        D[c] -= d\n\ndfs(0, -1, 0)\n   \nfor i in range(Q) :\n    print(ret[i])'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14339