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\n\ninput=sys.stdin.readline\nsys.setrecursionlimit(10**7)\n\nN,Q=map(int,input().split())\nedge=[[] for i in range(N)]\nfor i in range(N-1):\n    a,b,c,d=map(int,input().split())\n    edge[a-1].append((b-1,c,d))\n    edge[b-1].append((a-1,c,d))\n\nprv=[-1]*N\ndepth=[0]*N\ndist=[0 for i in range(N)]\ndef dfs(v,pv):\n    prv[v]=pv\n    for nv,color,weight in edge[v]:\n        if nv!=pv:\n            depth[nv]=depth[v]+1\n            dist[nv]=dist[v]+weight\n            dfs(nv,v)\n\ndfs(0,-1)\n\nLV = (N-1).bit_length()\ndef construct(prv):\n    kprv = [prv]\n    S = prv\n    for k in range(LV):\n        T = [0]*N\n        for i in range(N):\n            if S[i] is None:\n                continue\n            T[i] = S[S[i]]\n        kprv.append(T)\n        S = T\n    return kprv\n\nkprv=construct(prv)\n\ndef lca(u, v):\n    dd = depth[v] - depth[u]\n    if dd < 0:\n        u, v = v, u\n        dd = -dd\n\n    # assert depth[u] <= depth[v]\n    for k in range(LV+1):\n        if dd & 1:\n            v = kprv[k][v]\n        dd >>= 1\n\n    # assert depth[u] == depth[v]\n    if u == v:\n        return u\n\n    for k in range(LV-1, -1, -1):\n        pu = kprv[k][u]; pv = kprv[k][v]\n        if pu != pv:\n            u = pu; v = pv\n\n    # assert kprv[0][u] == kprv[0][v]\n    return kprv[0][u]\n\nquery=[]\nq=[[] for i in range(N)]\ndata={}\nfor i in range(Q):\n    x,y,u,v=map(int,input().split())\n    LCA=lca(u-1,v-1)\n    query.append((x,y,u-1,v-1,LCA))\n    q[u-1].append(x)\n    q[v-1].append(x)\n    q[LCA].append(x)\n\nnum=[0 for i in range(N)]\ncdist=[0 for i in range(N)]\n\ndef dfsc(v,pv,color,weight):\n    pn=num[color]\n    num[color]+=1\n    pd=cdist[color]\n    cdist[color]+=weight\n    for x in q[v]:\n        data[(v,x)]=(0,0)\n        data[(v,x)]=(num[x],cdist[x])\n\n    for nv,ncolor,nweight in edge[v]:\n        if nv!=pv:\n            dfsc(nv,v,ncolor,nweight)\n\n    num[color]=pn\n    cdist[color]=pd\n\n\n\ndfsc(0,0,0,0)\n\nfor x,y,u,v,LCA in query:\n    d1=dist[LCA]-data[(LCA,x)][1]+data[(LCA,x)][0]*y\n    d2=dist[u]-data[(u,x)][1]+data[(u,x)][0]*y\n    d3=dist[v]-data[(v,x)][1]+data[(v,x)][0]*y\n    print(d2-d1+d3-d1)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14327