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 PH push
#define MP make_pair
#define PB push_back
#define fst first
#define snd second
#define FOR(i, x, y) for(int i = (x); i < (y); ++i)
#define REP(i, x, y) for(int i = (x); i <= (y); ++i)
typedef double db;
typedef long long ll;
typedef long double ldb;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9 + 7;
const int maxn = 1e5 + 5;
const int logn = 20;

class Edge{
public:
	int u, v, c, w;
}ed[maxn];

class Query{
public:
	int u, v, x, y;
}qry[maxn];

int n, q;
ll sum;
int dep[maxn];
ll totw[maxn], cnt[maxn], ans[maxn];
int par[maxn][logn];
vector<int> g[maxn];
vector<pii> qq[maxn];

void dfs(int u, int p){
	if(~p) dep[u] = dep[p] + 1;
	par[u][0] = p;
	FOR(i, 0, g[u].size()){
		Edge e = ed[g[u][i]];
		int v = ((e.u == u) ? (e.v) : (e.u));
		if(v == p) continue;
		dfs(v, u);
	}
	return;
}

void precalc(){
	dfs(0, -1);
	FOR(i, 1, logn) FOR(u, 0, n){
		if(~par[u][i - 1]) par[u][i] = par[par[u][i - 1]][i - 1];
		else par[u][i] = -1;
	}
	return;
}

int lca(int u, int v){
	if(dep[u] < dep[v]) swap(u, v);
	int d = dep[u] - dep[v];
	FOR(i, 0, logn) if(d & (1 << i)) u = par[u][i];
	if(u == v) return u;
	for(int i = logn - 1; i >= 0; --i){
		if(~par[u][i] && par[u][i] != par[v][i])
			u = par[u][i], v = par[v][i];
	}
	return par[u][0];
}

void calc(int u, int p){
	FOR(i, 0, g[u].size()){
		Edge e = ed[g[u][i]];
		int v = ((e.u == u) ? (e.v) : (e.u));
		if(v == p) continue;
		++cnt[e.c];
		totw[e.c] += e.w;
		sum += e.w;
		calc(v, u);
		--cnt[e.c];
		totw[e.c] -= e.w;
		sum -= e.w;
	}
	FOR(i, 0, qq[u].size()){
		int tp = qq[u][i].fst, id = qq[u][i].snd;
		ans[id] += tp * (sum - totw[qry[id].x] + cnt[qry[id].x] * qry[id].y);
	}
	return;
}

int main(){
	scanf("%d%d", &n, &q);
	FOR(i, 1, n){
		scanf("%d%d%d%d", &ed[i].u, &ed[i].v, &ed[i].c, &ed[i].w);
		--ed[i].u; --ed[i].v;
		g[ed[i].u].PB(i);
		g[ed[i].v].PB(i);
	}
	precalc();
	FOR(i, 0, q){
		scanf("%d%d%d%d", &qry[i].x, &qry[i].y, &qry[i].u, &qry[i].v);
		--qry[i].u; --qry[i].v;
		int r = lca(qry[i].u, qry[i].v);
		qq[qry[i].u].PB(MP(1, i));
		qq[qry[i].v].PB(MP(1, i));
		qq[r].PB(MP(-2, i));
	}
	calc(0, -1);
	FOR(i, 0, q) printf("%lld\n", ans[i]);
	return 0;
}