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 <iostream>
#include <vector>
#include <unordered_map>
#include <map>
#include <unordered_set>
#include <set>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <climits>
#include <queue>
using namespace std;

const int M = 1e5 + 5;

struct Edge {
  int v, c, d;
};

struct Query {
  int id, x, y, delta;
};

int n, m;
vector<Edge> g[M];
vector<Query> qr[M];
int fa[M][19], dep[M], dis[M], ans[M], cnt[M], len[M];

void Link(int u, int v, int c, int d) {
  g[u].push_back({v, c, d});
}

void Dfs(int u) {
  for (auto &e : g[u]) {
    int v = e.v;
    if (v == fa[u][0]) continue;
    dep[v] = dep[u] + 1;
    fa[v][0] = u;
    for (int i = 1; i < 19; i++) {
      fa[v][i] = fa[fa[v][i - 1]][i - 1];
    }
    dis[v] = dis[u] + e.d;
    Dfs(v);
  }
}

int LCA(int u, int v) {
  if (dep[u] > dep[v]) swap(u, v);
  int diff = dep[v] - dep[u];
  for (int i = 0; i < 19; i++) {
    if (diff >> i & 1) {
      v = fa[v][i];
    }
  }
  if (u == v) return u;
  for (int i = 18; i >= 0; i--) {
    if (fa[u][i] != fa[v][i]) {
      u = fa[u][i];
      v = fa[v][i];
    }
  }
  return fa[u][0];
}

void Dfs2(int u) {
  for (auto &q : qr[u]) {
    ans[q.id] += (dis[u] - len[q.x] + cnt[q.x] * q.y) * q.delta;
  }

  for (auto &e : g[u]) {
    int v = e.v;
    if (v == fa[u][0]) continue;
    int c = e.c, d = e.d;
    cnt[c]++;
    len[c] += d;
    Dfs2(v);
    cnt[c]--;
    len[c] -= d;
  }
}

int main() {
  cin >> n >> m;

  for (int i = 1; i < n; i++) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    Link(a, b, c, d);
    Link(b, a, c, d);
  }

  Dfs(1);

  for (int i = 1; i <= m; i++) {
    int x, y, u, v;
    scanf("%d %d %d %d", &x, &y, &u, &v);
    int lca = LCA(u, v);
    qr[u].push_back({i, x, y, 1});
    qr[v].push_back({i, x, y, 1});
    qr[lca].push_back({i, x, y, -2});
  }

  Dfs2(1);

  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}