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>

#define fi first
#define se second
#define pii pair<int,int>
using namespace std;
const int maxn = 2e5+15;
int n,Q,a,b,c,d,cnt,head[maxn],ans[maxn];
struct edge{
    int to,nxt,col,len;
}e[maxn<<1];
struct query{
    int l,r,x,y,id;
}q[maxn<<1];
pii p[maxn];
inline void add(int u,int v,int c,int d){
    e[++cnt] = (edge){v,head[u],c,d};
    head[u] = cnt;
}
int dp[maxn][20],deep[maxn],st[maxn],ed[maxn],tot,A[maxn<<1];
void dfs(int x,int fa,int depth){
    A[++tot] = x; st[x] = tot; deep[x] = depth;
    for(int i = head[x];i > 0;i = e[i].nxt){
        int v = e[i].to;
        if(v == fa) continue;
        p[v] = pii(e[i].col,e[i].len);
        dp[v][0] = x; dfs(v,x,depth+1);
    }
    A[++tot] = x; ed[x] = tot;
}
void init(){
    for(int i = 1;(1<<i)<=n; ++i)
    for(int j = 1;j <= n; ++j)
    dp[j][i] = dp[dp[j][i-1]][i-1];
}
int LCA(int u,int v){
    if(deep[u] < deep[v]) swap(u,v);
    int h = deep[u] - deep[v];
    for(int i = 0;i < 20; ++i){
        if(h&(1<<i)) u = dp[u][i];
    }
    if(u == v) return u;
    for(int i = 19; ~i; --i){
        if(dp[u][i] != dp[v][i]){
            u = dp[u][i];
            v = dp[v][i];
        }
    }
    return dp[u][0];
}
int block,num[maxn],sum[maxn],vis[maxn],res;
bool cmp(query a,query b){
    return (a.l/block)^(b.l/block)?a.l<b.l:(((a.l/block)&1)?a.r<b.r:a.r>b.r);
}
void add(int v){
    if(vis[v]) {num[p[v].fi]--;sum[p[v].fi]-=p[v].se;res-=p[v].se;}
    else {num[p[v].fi]++;sum[p[v].fi]+=p[v].se;res+=p[v].se;}
    vis[v] ^= 1;
}
void del(int v){
    if(vis[v]) {num[p[v].fi]--;sum[p[v].fi]-=p[v].se;res-=p[v].se;}
    else {num[p[v].fi]++;sum[p[v].fi]+=p[v].se;res+=p[v].se;}
    vis[v] ^= 1;
}
int main(){
    scanf("%d %d",&n,&Q); block = sqrt(n*1.0);
    for(int i = 1;i < n; ++i){
        scanf("%d %d %d %d",&a,&b,&c,&d);
        add(a,b,c,d); add(b,a,c,d);
    }
    dfs(1,0,1); init();
    for(int i = 0;i < Q; ++i){
        scanf("%d %d %d %d",&a,&b,&c,&d);
        if(st[c] > st[d]) swap(c,d);
        if(LCA(c,d) == c) q[i] = (query){st[c]+1,st[d],a,b,i};
        else q[i] = (query){ed[c],st[d],a,b,i}; 
    }
    sort(q,q+Q,cmp);
    int L = 2,R = 1; res = 0;
    for(int i = 0;i < Q; ++i){
        while(R > q[i].r) del(A[R]),--R;
        while(R < q[i].r) ++R,add(A[R]);
        while(L < q[i].l) del(A[L]),++L;
        while(L > q[i].l) --L,add(A[L]);
        ans[q[i].id] = res-sum[q[i].x]+num[q[i].x]*q[i].y;
    }
    for(int i = 0;i < Q; ++i) printf("%d\n",ans[i]);
    return 0;
}