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;
#define ll long long

int N, Q;
vector<vector<pair<int, int>>> edge;
vector<vector<ll>> tour_cnt, tour_num, tour_sum;
vector<ll> sum;
vector<ll> num;
vector<ll> arrive;
vector<ll> depth;
vector<vector<int>> parent;
vector<ll> dist;
vector<ll> A, B, C, D;
ll d = 0;
int cnt = 0;
int dp = 0;

void dfs(int now, int pre) {
    depth[now] = dp;
    arrive[now] = cnt;
    dist[now] = d;
    for (auto p : edge[now]) {
        int nxt = p.first;
        int i = p.second;
        if (pre == nxt) continue;
        parent[nxt][0] = now;
        d += D[i];
        sum[C[i]] += D[i];
        num[C[i]]++;
        cnt++;
        tour_cnt[C[i]].push_back(cnt);
        tour_sum[C[i]].push_back(sum[C[i]]);
        tour_num[C[i]].push_back(num[C[i]]);
        
        dp++;
        dfs(nxt, now);
        dp--;
        
        d -= D[i];
        sum[C[i]] -= D[i];
        num[C[i]]--;
        
        cnt++;
        tour_cnt[C[i]].push_back(cnt);
        tour_sum[C[i]].push_back(sum[C[i]]);
        tour_num[C[i]].push_back(num[C[i]]);
    }
    return;
}

int main() {
    cin >> N >> Q;
    A.resize(N-1);
    B.resize(N-1);
    C.resize(N-1);
    D.resize(N-1);
    edge.resize(N+1);
    parent = vector<vector<int>>(N+1, vector<int>(20, 0));
    dist.resize(N+1, 0);
    tour_cnt.resize(N);
    tour_num.resize(N);
    tour_sum.resize(N);
    sum.resize(N, 0);
    num.resize(N, 0);
    arrive.resize(N+1, 0);
    depth.resize(N+1, 0);
    for (int i = 0; i < N-1; i++) {
        cin >> A[i] >> B[i] >> C[i] >> D[i];
        edge[A[i]].push_back({B[i], i});
        edge[B[i]].push_back({A[i], i});
    }
    
    for (int i = 1; i < N; i++) {
        tour_cnt[i].push_back(-1);
        tour_num[i].push_back(0);
        tour_sum[i].push_back(0);
    }
    
    dfs(1, 0);
    for (int i = 1; i < N; i++) {
        tour_cnt[i].push_back(2e9);
        tour_num[i].push_back(2e9);
        tour_sum[i].push_back(2e9);
    }
    
    
    for (int i = 1; i <= 19; i++) {
        for (int j = 1; j <= N; j++) {
            parent[j][i] = parent[parent[j][i-1]][i-1];
        }
    }
    
    for (int z = 0; z < Q; z++) {
        ll x, y, u, v;
        cin >> x >> y >> u >> v;
        if (depth[u] > depth[v]) swap(u, v);
        int b = depth[v]-depth[u];
        int ua = u, va = v, a;
        for (int i = 0; i < 19; i++) {
            if ((b>>i)&1) va = parent[va][i];
        }
        a = va;
        
        while (ua != va) {
            if (parent[ua][0] == parent[ua][0]) {
                a = parent[ua][0];
                break;
            }
            int t = 0;
            while (parent[ua][t+1] != parent[va][t+1]) t++;
            ua = parent[ua][t];
            va = parent[va][t];
            a = va;
        }
        
      
        ll ans = dist[u]+dist[v]-dist[a]*2;
        int g;
        auto it = lower_bound(tour_cnt[x].begin(), tour_cnt[x].end(), arrive[u]);
        
        
        if (*it == arrive[u]) {
            g = distance(tour_cnt[x].begin(), it);
        }else {
            g = distance(tour_cnt[x].begin(), --it);
        }
        ans -= tour_sum[x][g];
        ans += y * tour_num[x][g];
        
        
        it = lower_bound(tour_cnt[x].begin(), tour_cnt[x].end(), arrive[v]);
        if (*it == arrive[v]) {
            g = distance(tour_cnt[x].begin(), it);
        }else {
            g = distance(tour_cnt[x].begin(), --it);
        }
        ans -= tour_sum[x][g];
        ans += y * tour_num[x][g];
        
        
        
        it = lower_bound(tour_cnt[x].begin(), tour_cnt[x].end(), arrive[a]);
        if (*it == arrive[a]) {
            g = distance(tour_cnt[x].begin(), it);
        }else {
            g = distance(tour_cnt[x].begin(), --it);
        }
        ans += tour_sum[x][g]*2;
        ans -= y * tour_num[x][g]*2;
        
        cout << ans << endl;
    }
    
    return 0;
}