Problem: 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.

Example

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


Output

130
200
60

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;

struct edge {
  int to, w, c;
};
struct query {
  int u, v, x, y;
};
vector<edge> adj[maxn];
query que[maxn];
vector<int> qid[maxn];
int sum[maxn], depth[maxn], parent[maxn][30], n, dist, ncolor[maxn], sumco[maxn];
unordered_map<int, int> ccnt[maxn], csum[maxn];

void dfs(int u, int fa, int d) {
  depth[u] = d;
  parent[u][0] = fa;
  for (auto e : adj[u]) {
    if (e.to != fa) dfs(e.to, u, d + 1);
  }
}

void init() {
  for (int i = 1; i <= 20; ++i) {
    for (int j = 0; j < n; ++j) {
      if (parent[j][i - 1] < 0) parent[j][i] = -1;
      else parent[j][i] = parent[parent[j][i - 1]][i - 1];
    }
  }
}

int lca(int u, int v) {
  if (depth[u] > depth[v]) swap(u, v);
  for (int i = 20; i >= 0; --i)
    if ((depth[v] - depth[u]) >> i & 1) v = parent[v][i];
  if (u == v) return u;
  for (int i = 20; i >= 0; --i) {
    if (parent[u][i] != parent[v][i]) {
      v = parent[v][i];
      u = parent[u][i];
    }
  }
  return parent[u][0];
}

void dfs2(int u, int p) {
  sum[u] = dist;
  for (auto q : qid[u]) {
    ccnt[u][que[q].x] = ncolor[que[q].x];
    csum[u][que[q].x] = sumco[que[q].x];
  }
  for (auto e : adj[u]) {
    if (e.to != p) {
      dist += e.w;
      ++ncolor[e.c];
      sumco[e.c] += e.w;
      dfs2(e.to, u);
      dist -= e.w;
      --ncolor[e.c];
      sumco[e.c] -= e.w;
    }
  }
}

int main() {
  #ifdef LOCAL
  freopen("input.txt", "r", stdin);
  #endif
  ios::sync_with_stdio(false);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v, w, c;
    cin >> u >> v >> c >> w;
    --u; --v;
    adj[u].push_back((edge){v, w, c});
    adj[v].push_back((edge){u, w, c});
  }
  dfs(0, -1, 0);
  init();
  for (int i = 0; i < q; ++i) {
    int u, v, x, y;
    cin >> x >> y >> u >> v;
    --u; --v;
    que[i] = (query){u, v, x, y};
    qid[lca(u, v)].push_back(i);
    qid[u].push_back(i);
    qid[v].push_back(i);
  }
  dfs2(0, -1);
  for (int i = 0; i < q; ++i) {
    int t1 = sum[que[i].u] - csum[que[i].u][que[i].x] + ccnt[que[i].u][que[i].x] * que[i].y;
    int t2 = sum[que[i].v] - csum[que[i].v][que[i].x] + ccnt[que[i].v][que[i].x] * que[i].y;
    int l = lca(que[i].u, que[i].v);
    int t3 = sum[l] - csum[l][que[i].x] + ccnt[l][que[i].x] * que[i].y;
    cout << t1 + t2 - 2 * t3 << endl;
  }
  return 0;
}