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;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb emplace_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define add(x) (x > MOD ? x-MOD :(x<0 ? x+MOD:x))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1e5 + 5;
vector<tuple<int, int, int>> g[N];
vector<tuple<int, int, int, int>> vc[N];
int n, m, a, b, c, d, x, y, u, v;
int anc[N][18], deep[N], len[N], ans[N];
int cur_cnt[N], cur_sum[N];
void dfs(int u, int o) {
  deep[u] = deep[o] + 1;
  anc[u][0] = o;
  for (int i = 1; i < 18; ++i)
    anc[u][i] = anc[anc[u][i - 1]][i - 1];
  for (auto t : g[u]) {
    int v = get<0>(t), w = get<2>(t);
    if (v != o) {
      len[v] = len[u] + w;
      dfs(v, u);
    }
  }
}
void DFS(int u, int o) {
  for (int i = 0; i < vc[u].size(); ++i) {
    auto t = vc[u][i];
    int id = get<0>(t), c = get<1>(t), w = get<2>(t), f = get<3>(t);
    ans[id] -= f * cur_sum[c];
    ans[id] += f * cur_cnt[c] * w;
  }
  for (auto t : g[u]) {
    int v = get<0>(t), c = get<1>(t), w = get<2>(t);
    if (v != o) {
      cur_cnt[c]++;
      cur_sum[c] += w;
      DFS(v, u);
      cur_cnt[c]--;
      cur_sum[c] -= w;
    }
  }
}
inline int lca(int u, int v) {
  if (deep[u] < deep[v])
    swap(u, v);
  for (int i = 17; i >= 0; --i)
    if (deep[anc[u][i]] >= deep[v])
      u = anc[u][i];
  if (u == v)
    return u;
  for (int i = 17; i >= 0; --i)
    if (anc[u][i] != anc[v][i])
      u = anc[u][i], v = anc[v][i];
  return anc[u][0];
}
int main() {
//  freopen("input.in", "rt", stdin);
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; ++i)
    scanf("%d %d %d %d", &a, &b, &c, &d), g[a].pb(b, c, d), g[b].pb(a, c, d);
  dfs(1, 1);
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d %d", &x, &y, &u, &v);
    int a = lca(u, v);
    ans[i] = len[u] + len[v] - 2 * len[a];
    vc[u].pb(i, x, y, 1);
    vc[v].pb(i, x, y, 1);
    vc[a].pb(i, x, y, -2);
  }
  DFS(1, 1);
  for (int i = 1; i <= m; ++i)
    printf("%d\n", ans[i]);
  return 0;
}