Problem:
Score : 600 points

Problem Statement

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 1

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 1

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.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
#define rep(i, a, b) for (int i = (a); i < (b); ++i)

#pragma region Graph Template
template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;
#pragma endregion
#pragma region Doubling-LCA
template< typename G >
struct DoublingLowestCommonAncestor {
  const int LOG;
  vector< int > dep;
  const G &g;
  vector< vector< int > > table;

  DoublingLowestCommonAncestor(const G &g) : g(g), dep(g.size()), LOG(32 - __builtin_clz(g.size())) {
    table.assign(LOG, vector< int >(g.size(), -1));
  }

  void dfs(int idx, int par, int d) {
    table[0][idx] = par;
    dep[idx] = d;
    for(auto &to : g[idx]) {
      if(to != par) dfs(to, idx, d + 1);
    }
  }

  void build() {
    dfs(0, -1, 0);
    for(int k = 0; k + 1 < LOG; k++) {
      for(int i = 0; i < table[k].size(); i++) {
        if(table[k][i] == -1) table[k + 1][i] = -1;
        else table[k + 1][i] = table[k][table[k][i]];
      }
    }
  }

  int query(int u, int v) {
    if(dep[u] > dep[v]) swap(u, v);
    for(int i = LOG - 1; i >= 0; i--) {
      if(((dep[v] - dep[u]) >> i) & 1) v = table[i][v];
    }
    if(u == v) return u;
    for(int i = LOG - 1; i >= 0; i--) {
      if(table[i][u] != table[i][v]) {
        u = table[i][u];
        v = table[i][v];
      }
    }
    return table[0][u];
  }
};
#pragma endregion

const int MX = 100010;
int N, Q;
UnWeightedGraph g;
WeightedGraph<int> wg;

int dist_from_root[MX];
void build_dst(int cur, int prv = -1, int d = 0) {
    dist_from_root[cur] = d;
    for (auto& nxt: wg[cur]) {
        if (nxt.to == prv) continue;
        build_dst(nxt.to, cur, d + nxt.cost);
    } 
}

int cnt[MX], clr[MX], dst[MX];
vector<int> need[MX];

// [clr, idx] -> ans
map<pair<int, int>, int> clrcnt;
map<pair<int, int>, int> clrdst;

int clrcnt_tmp[MX];
int clrdst_tmp[MX];
void euler_tour(int cur, int prv = -1) {
    clrcnt_tmp[clr[cur]]++;
    clrdst_tmp[clr[cur]] += dst[cur];
    for (auto& n: need[cur]) {
        clrcnt[make_pair(n, cur)] = clrcnt_tmp[n];
        clrdst[make_pair(n, cur)] = clrdst_tmp[n];
    }
    for (auto& nxt: g[cur]) if (prv != nxt) euler_tour(nxt, cur);
    clrcnt_tmp[clr[cur]]--;
    clrdst_tmp[clr[cur]] -= dst[cur];
}

signed main() {

    cin >> N >> Q;
    g.resize(N);
    wg.resize(N);

    vector<tuple<int, int, int, int>> tmp;
    rep(i, 0, N - 1) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        --a, --b, --c;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
        wg[a].emplace_back(b, d);
        wg[b].emplace_back(a, d);
        tmp.emplace_back(a, b, c, d);
    }
    DoublingLowestCommonAncestor<UnWeightedGraph> lca(g);
    lca.build();

    build_dst(0);

    clr[0] = -1;
    for (auto& t: tmp) {
        int a, b, c, d;
        tie(a, b, c, d) = t;
        int dep = lca.query(a, b) ^ a ^ b;
        clr[dep] = c;
        dst[dep] = d;
    }

    vector<tuple<int, int, int, int>> queries;
    rep(i, 0, Q) {
        int x, y, u, v;
        cin >> x >> y >> u >> v;
        --x, --u, --v;
        queries.emplace_back(x, y, u, v);
        need[u].emplace_back(x);
        need[v].emplace_back(x);
        need[lca.query(u, v)].emplace_back(x);
    }

    euler_tour(0);

    for (auto& q: queries) {
        int x, y, u, v;
        tie(x, y, u, v) = q;
        int l = lca.query(u, v);
        lint ans = dist_from_root[u] + dist_from_root[v] - dist_from_root[l] * 2;
        ans -= clrdst[make_pair(x, u)] + clrdst[make_pair(x, v)] - clrdst[make_pair(x, l)] * 2;
        ans += (clrcnt[make_pair(x, u)] + clrcnt[make_pair(x, v)] - clrcnt[make_pair(x, l)] * 2) * y;
        cout << ans << endl;
    }

}