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(10**7)\n\nN, Q = list(map(int, input().split()))\nedge = [[] for _ in range(N+1)]\nfor _ in range(N-1):\n        a, b, c, d = list(map(int, input().split()))\n        edge[a].append((b, c, d))\n        edge[b].append((a, c, d))\n        \ndepth = [0] * (N+1)\nU = 17\nancestor = [[0] * (U+1) for _ in range(N+1)]\n\nq = [(1, 0, 0)]\nwhile q:\n        qq = []\n        for x, d, parent in q:\n                depth[x] = d\n                ax = ancestor[x]\n                ax[0] = parent\n                for i in range(d.bit_length()-1):\n                        ax[i+1] = ancestor[ax[i]][i]\n\n                for y, _, _ in edge[x]:\n                        if y == parent:\n                                continue\n                        qq.append((y, d+1, x))\n        q = qq\n\ndef lca(x, y):\n        dx = depth[x]\n        dy = depth[y]\n        if dx > dy:\n                x, y = y, x\n                dx, dy = dy, dx\n        diff = dy - dx\n        while diff:\n                step = diff & (-diff)\n                y = ancestor[y][step.bit_length()-1]\n                diff -= step\n        while x!=y:\n                j = 0\n                while ancestor[x][j] != ancestor[y][j]:\n                        j += 1\n                if j==0:\n                        return ancestor[x][0]\n                x = ancestor[x][j-1]\n                y = ancestor[y][j-1]\n        return x\n\ntask = [[] for _ in range(N+1)]\nfor i in range(Q):\n        x, y, u, v = list(map(int, input().split()))\n        task[u].append((i, x, y, 1))\n        task[v].append((i, x, y, 1))\n        task[lca(u, v)].append((i, x, y, -2))\n\nanswer = [0] * Q\n\ndef dfs(x=1, sums=[0]*(N+1), nums=[0]*(N+1), total=0, parent=0):\n        for i, c, d, coef in task[x]:\n                answer[i] += coef * (total - sums[c] + nums[c]*d)\n        for y, c, d in edge[x]:\n                if y == parent:\n                        continue\n                sums[c] += d\n                nums[c] += 1\n                total += d\n                dfs(y, sums, nums, total, x)\n                sums[c] -= d\n                nums[c] -= 1\n                total -= d\n        return\n\ndfs()\n\nprint(('\\n'.join(map(str, answer))))\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14101