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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
// #define int ll
using PII = pair<ll, ll>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()

template<typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }

template<typename T> vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts) {
    return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type
fill_v(T &t, const V &v) { t=v; }
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t, const V &v ) { for(auto &e:t) fill_v(e,v); }

template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a) {
    out<<'('<<a.first<<','<<a.second<<')'; return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
    out<<'[';
    for(const T &i: a) out<<i<<',';
    out<<']';
    return out;
}
template<class T>
ostream &operator <<(ostream& out, const set<T>& a) {
    out<<'{';
    for(const T &i: a) out<<i<<',';
    out<<'}';
    return out;
}
template<class T, class S>
ostream &operator <<(ostream& out, const map<T,S>& a) {
    out<<'{';
    for(auto &i: a) out<<i<<',';
    out<<'}';
    return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; // DRUL
const int INF = 1<<30;
const ll LLINF = 1LL<<60;
const ll MOD = 1000000007;

struct LCA {
    const int n = 0;
    const int log2_n = 0;
    vector<vector<int>> g;
    vector<vector<int>> par;  // par[2^i上][頂点v]
    vector<int> dep;

    void dfs(int v, int p, int d) {
        par[0][v] = p;
        dep[v] = d;
        for(auto to: g[v]) {
            if(to == p) continue;
            dfs(to, v, d+1);
        }
    }

    LCA() {}
    LCA(int n) : n(n), log2_n(log2(n)+1), g(n),
        par(log2_n, vector<int>(n)), dep(n) {}

    void add_edge(int u, int v) {
        g[u].push_back(v);
        g[v].push_back(u);
    }

    void build() {
        dfs(/*root=*/0, -1, 0);
        for(int i=0; i+1 < log2_n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(par[i][j] < 0) par[i+1][j] = -1;
                else par[i+1][j] = par[i][par[i][j]];
            }
        }
    }

    int get(int u, int v) {
        if(dep[u] > dep[v]) swap(u, v);
        REP(i, log2_n) {
            if((dep[v] - dep[u]) >> i & 1) {
                v = par[i][v];
            }
        }
        if(u == v) return u;
        for(int i=log2_n-1; i>=0; --i) {
            if(par[i][u] != par[i][v]) {
                u = par[i][u];
                v = par[i][v];
            }
        }
        return par[0][u];
    }
};

signed main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n, q;
    cin >> n >> q;
    LCA graph(n);
    struct edge { ll to, col, dist; };
    vector<vector<edge>> g(n);
    REP(i, n-1) {
        ll a, b, c, d;
        cin >> a >> b >> c >> d;
        a--, b--, c--;
        g[a].push_back({b, c, d});
        g[b].push_back({a, c, d});
        graph.add_edge(a, b);
    }
    graph.build();
    vector<ll> x(q), y(q);
    vector<vector<pair<ll,bool>>> query(n);
    REP(i, q) {
        ll u, v, lca;
        cin >> x[i] >> y[i] >> u >> v;
        x[i]--, u--, v--;
        lca = graph.get(u, v);
        query[u].push_back({i, true});
        query[v].push_back({i, true});
        query[lca].push_back({i, false});
    }

    vector<ll> ans(q);
    function<void(ll,ll,ll,map<ll,PII>&)> dfs = [&](ll v, ll p, ll sum, map<ll,PII> &mp) {
        for(auto p: query[v]) {
            if(p.second) ans[p.first] += sum - mp[x[p.first]].second + mp[x[p.first]].first * y[p.first];
            else ans[p.first] -= sum - mp[x[p.first]].second + mp[x[p.first]].first * y[p.first];
        }
        for(auto to: g[v]) {
            if(to.to == p) continue;
            if(mp.find(to.col) == mp.end()) {
                mp[to.col] = PII(1, to.dist);
            } else {
                mp[to.col] = PII(mp[to.col].first+1, mp[to.col].second+to.dist);
            }
            dfs(to.to, v, sum + to.dist, mp);
            mp[to.col] = PII(mp[to.col].first-1, mp[to.col].second-to.dist);
            if(mp[to.col].first == 0) {
                mp.erase(to.col);
            }
        }
    };

    map<ll,PII> mp;
    dfs(0, -1, 0, mp);
    for(auto i: ans) cout << i << endl;

    return 0;
}