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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <tuple>
using namespace std;

int n, q, ans[100005], dis[100005];
vector <tuple <int, int, int> > g[100005];

int par[100005], sz[100005];
int son[100005], dep[100005];
int ind[100005], top[100005], indx;
void dfs1(int u, int parent, int depth) {
  dep[u] = depth, par[u] = parent, sz[u] = 1;
  for (int i = 0; i < (int)g[u].size(); ++ i) {
    int v = get <0> (g[u][i]);
    if (v == parent) continue;
    dfs1(v, u, depth + 1), sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int topv) {
  top[u] = topv, ind[u] = ++ indx;
  if (son[u]) dfs2(son[u], topv);
  for (int i = 0; i < (int)g[u].size(); ++ i) {
    int v = get <0> (g[u][i]);
    if (v == par[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
inline int get_lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]])
      swap(u, v);
    u = par[top[u]];
  }
  if (dep[u] < dep[v]) return u;
  return v;
}

int dist[100005];
void dfs3(int u, int parent) {
  for (int i = 0; i < (int)g[u].size(); ++ i) {
    int v = get <0>(g[u][i]), w = get <2>(g[u][i]);
    if (v == parent) continue;
    dist[v] = dist[u] + w, dfs3(v, u);
  }
}

vector <tuple <int, int, int> > qry[100005];
int d[100005], cnt[100005];
void dfs4(int u, int parent) {
  for (int i = 0; i < (int)g[u].size(); ++ i) {
    int v = get <0>(g[u][i]), c = get <1> (g[u][i]), w = get <2>(g[u][i]);
    if (v == parent) continue;
    d[c] += w, cnt[c] ++;
    dfs4(v, u);
    d[c] -= w, cnt[c] --;
  }
  for (int i = 0; i < (int)qry[u].size(); ++ i) {
    int id = get <0>(qry[u][i]), c = get <1>(qry[u][i]);
    int t = get <2>(qry[u][i]), nowd = dis[id];
    int x = d[c], y = nowd * cnt[c];
    ans[id] += (y - x) * t;
  }
}

int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i < n; ++ i) {
    int u, v, c, d;
    scanf("%d %d %d %d", &u, &v, &c, &d);
    g[u].emplace_back(v, c, d);
    g[v].emplace_back(u, c, d);
  }
  dfs1(1, 0, 1);
  dfs2(1, 1);
  dfs3(1, 0);
  for (int i = 1; i <= q; ++ i) {
    int c, u, v;
    scanf("%d %d %d %d", &c, dis + i, &u, &v);
    int lca = get_lca(u, v);
    ans[i] = dist[u] + dist[v] - 2 * dist[lca];
    qry[u].emplace_back(i, c, 1);
    qry[v].emplace_back(i, c, 1);
    qry[lca].emplace_back(i, c, -2);
  }
  dfs4(1, 0);
  for (int i = 1; i <= q; ++ i)
    printf("%d\n", ans[i]);
  return 0;
}