#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100, lg = 20;
struct query {
	int i, x, y, sig;
};
vector<tuple<int,int,int> > e[maxn];
vector<query> que[maxn];
int par[lg][maxn], dep[maxn];
void dfs(int u, int p){
	par[0][u] = p;
	for(auto ee : e[u]){
		int to, b, c; tie(to, b, c) = ee;
		if(to != p){
			dep[to] = dep[u] + 1;
			dfs(to, u);
		}
	}
}
int lca(int u, int v){
	if(dep[u] < dep[v]) swap(u, v);
	for(int i = lg - 1; i >= 0; i--) if((dep[u] - dep[v])&(1<<i)) u = par[i][u];
	if(u == v) return u;
	for(int i = lg - 1; i >= 0; i--) if(par[i][u] != par[i][v]) u = par[i][u], v = par[i][v];
	return par[0][u];
}
int ans[maxn], tot[maxn], amn[maxn];
void solve(int u, int p, int d){
	for(auto q : que[u]){
		ans[q.i] += q.sig * (d - tot[q.x] + q.y * amn[q.x]);
	}
	for(auto ee : e[u]){
		int to, c, di; tie(to, c, di) = ee;
		if(to == p) continue;
		tot[c] += di;
		amn[c] += 1;
		solve(to, u, d + di);
		tot[c] -= di;
		amn[c] -= 1;
	}
}
int main(){
	int n, q; cin >> n >> q;
	for(int i = 0; i < n - 1; i++){
		int a, b, c, d; cin >> a >> b >> c >> d;
		e[a].push_back(make_tuple(b, c, d));
		e[b].push_back(make_tuple(a, c, d));
	}
	dfs(1, 0);
	for(int i = 1; i < lg; i++){
		for(int j = 1; j <= n; j++)
			if(par[i - 1][j])
				par[i][j] = par[i - 1][par[i - 1][j]];
	}
	for(int i = 0; i < q; i++){
		int x, y, u, v; cin >> x >> y >> u >> v;
		que[u].push_back({i, x, y, 1});
		que[v].push_back({i, x, y, 1});
		que[lca(u, v)].push_back({i, x, y, -2});
	}
	solve(1, 0, 0);
	for(int i = 0; i < q; i++) cout << ans[i] << endl;
	return 0;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100, lg = 20;
struct query {
	int i, x, y, sig;
};
vector<tuple<int,int,int> > e[maxn];
vector<query> que[maxn];
int par[lg][maxn], dep[maxn];
void dfs(int u, int p){
	par[0][u] = p;
	for(auto ee : e[u]){
		int to, b, c; tie(to, b, c) = ee;
		if(to != p){
			dep[to] = dep[u] + 1;
			dfs(to, u);
		}
	}
}
int lca(int u, int v){
	if(dep[u] < dep[v]) swap(u, v);
	for(int i = lg - 1; i >= 0; i--) if((dep[u] - dep[v])&(1<<i)) u = par[i][u];
	if(u == v) return u;
	for(int i = lg - 1; i >= 0; i--) if(par[i][u] != par[i][v]) u = par[i][u], v = par[i][v];
	return par[0][u];
}
int ans[maxn], tot[maxn], amn[maxn];
void solve(int u, int p, int d){
	for(auto q : que[u]){
		ans[q.i] += q.sig * (d - tot[q.x] + q.y * amn[q.x]);
	}
	for(auto ee : e[u]){
		int to, c, di; tie(to, c, di) = ee;
		if(to == p) continue;
		tot[c] += di;
		amn[c] += 1;
		solve(to, u, d + di);
		tot[c] -= di;
		amn[c] -= 1;
	}
}
int main(){
	int n, q; cin >> n >> q;
	for(int i = 0; i < n - 1; i++){
		int a, b, c, d; cin >> a >> b >> c >> d;
		e[a].push_back(make_tuple(b, c, d));
		e[b].push_back(make_tuple(a, c, d));
	}
	dfs(1, 0);
	for(int i = 1; i < lg; i++){
		for(int j = 1; j <= n; j++)
			if(par[i - 1][j])
				par[i][j] = par[i - 1][par[i - 1][j]];
	}
	for(int i = 0; i < q; i++){
		int x, y, u, v; cin >> x >> y >> u >> v;
		que[u].push_back({i, x, y, 1});
		que[v].push_back({i, x, y, 1});
		que[lca(u, v)].push_back({i, x, y, -2});
	}
	solve(1, 0, 0);
	for(int i = 0; i < q; i++) cout << ans[i] << endl;
	return 0;
}


```