#include<iostream>
#include<vector>
using namespace std;
const int maxn = 100001;
const int maxk = 17;
#define is_anc(a, b) (in[a] < in[b] and out[a] >= out[b])
struct info{
	int v, c, w;
};
struct query{
	int id, c, w, y;
};
int n, anc[maxn][maxk], in[maxn], out[maxn], depth[maxn], m;
long long ans[maxn];
vector < info > edge[maxn];
vector < query > ask[maxn];
void dfs(int now, int last = 1){
	static int t = 0;
	static int d = 0;
	depth[now] = d++;
	in[now] = t++;
	anc[now][0] = last;
	for(int i = 1;i < maxk;++i)anc[now][i] = anc[anc[now][i - 1]][i - 1];
	for(auto &i : edge[now])
		if(last != i.v)dfs(i.v, now);
	out[now] = t;
	--d;
}
int get_lca(int a, int b){
	if(depth[a] > depth[b])swap(a, b);
	if(is_anc(a, b))return a;
	for(int i = maxk - 1;i > -1;--i)
		if(!is_anc(anc[b][i], a))b = anc[b][i];
	return anc[b][0];
}
void dfs2(int now, int last = 1){
	static int total_w = 0;
	static int q_of[maxn] = {};
	static int w_of[maxn] = {};
	for(auto &i : ask[now]){
		ans[i.id] += (total_w - w_of[i.c] + q_of[i.c] * i.y) * i.w;
	}
	for(auto &i : edge[now]){
		if(i.v != last){
			++q_of[i.c];
			w_of[i.c] += i.w;
			total_w += i.w;
			dfs2(i.v, now);
			--q_of[i.c];
			w_of[i.c] -= i.w;
			total_w -= i.w;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m;
	for(int a, b, c, d, i = 1;i < n;++i){
		cin >> a >> b >> c >> d;
		edge[a].push_back({b, c, d});
		edge[b].push_back({a, c, d});
	}
	dfs(1);
	for(int i = 0, x, y, u, v, lca;i < m;++i){
        cin >> x >> y >> u >> v;
		lca = get_lca(u, v);
		ask[u].push_back({i, x, 1, y});
		ask[v].push_back({i, x, 1, y});
		ask[lca].push_back({i, x, -2, y});
	}
	dfs2(1);
	for(int i = 0;i < m;++i)cout << ans[i] << '\n';
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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<iostream>
#include<vector>
using namespace std;
const int maxn = 100001;
const int maxk = 17;
#define is_anc(a, b) (in[a] < in[b] and out[a] >= out[b])
struct info{
	int v, c, w;
};
struct query{
	int id, c, w, y;
};
int n, anc[maxn][maxk], in[maxn], out[maxn], depth[maxn], m;
long long ans[maxn];
vector < info > edge[maxn];
vector < query > ask[maxn];
void dfs(int now, int last = 1){
	static int t = 0;
	static int d = 0;
	depth[now] = d++;
	in[now] = t++;
	anc[now][0] = last;
	for(int i = 1;i < maxk;++i)anc[now][i] = anc[anc[now][i - 1]][i - 1];
	for(auto &i : edge[now])
		if(last != i.v)dfs(i.v, now);
	out[now] = t;
	--d;
}
int get_lca(int a, int b){
	if(depth[a] > depth[b])swap(a, b);
	if(is_anc(a, b))return a;
	for(int i = maxk - 1;i > -1;--i)
		if(!is_anc(anc[b][i], a))b = anc[b][i];
	return anc[b][0];
}
void dfs2(int now, int last = 1){
	static int total_w = 0;
	static int q_of[maxn] = {};
	static int w_of[maxn] = {};
	for(auto &i : ask[now]){
		ans[i.id] += (total_w - w_of[i.c] + q_of[i.c] * i.y) * i.w;
	}
	for(auto &i : edge[now]){
		if(i.v != last){
			++q_of[i.c];
			w_of[i.c] += i.w;
			total_w += i.w;
			dfs2(i.v, now);
			--q_of[i.c];
			w_of[i.c] -= i.w;
			total_w -= i.w;
		}
	}
}
int main(){
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m;
	for(int a, b, c, d, i = 1;i < n;++i){
		cin >> a >> b >> c >> d;
		edge[a].push_back({b, c, d});
		edge[b].push_back({a, c, d});
	}
	dfs(1);
	for(int i = 0, x, y, u, v, lca;i < m;++i){
        cin >> x >> y >> u >> v;
		lca = get_lca(u, v);
		ask[u].push_back({i, x, 1, y});
		ask[v].push_back({i, x, 1, y});
		ask[lca].push_back({i, x, -2, y});
	}
	dfs2(1);
	for(int i = 0;i < m;++i)cout << ans[i] << '\n';
}

```