Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

130
200
60

The graph in this input is as follows:

Here the edges of Color 1 are shown as solid red lines, the edge of Color 2 is shown as a bold green line, and the edge of Color 4 is shown as a blue dashed line.

Query 1: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 4 is 100 + 30 = 130.

Query 2: Assuming that the length of every edge whose color is 1 is changed to 100, the distance between Vertex 1 and Vertex 5 is 100 + 100 = 200.

Query 3: Assuming that the length of every edge whose color is 3 is changed to 1000 (there is no such edge), the distance between Vertex 3 and Vertex 4 is 20 + 10 + 30 = 60. Note that the edges of Color 1 now have their original lengths.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

const B: usize = 17;
const W: usize = 100_000;

fn calc<T, U>(v: usize, p: usize, g: &[Vec<(usize, T, U)>],
              par: &mut [usize], dep: &mut [i32], d: i32) {
    dep[v] = d;
    let n = g.len();
    if p < n {
        par[v] = p;
    }
    for &(w, _, _) in &g[v] {
        if w == p { continue; }
        calc(w, v, g, par, dep, d + 1);
    }
}

fn dfs(v: usize, p: usize, g: &[Vec<(usize, usize, i64)>],
       qs: &[Vec<(usize, i64, i64, usize)>], ans: &mut [i64],
       freq: &mut [(i64, i64)], tot: &mut i64,
) {
    for &(col, changed, fac, idx) in &qs[v] {
        let (freq, sum) = freq[col];
        let tmp = *tot - sum + freq * changed;
        ans[idx] += fac * tmp;
    }
    for &(w, x, y) in &g[v] {
        if w == p { continue; }
        freq[x].0 += 1;
        freq[x].1 += y;
        *tot += y;
        dfs(w, v, g, qs, ans, freq, tot);
        freq[x].0 -= 1;
        freq[x].1 -= y;
        *tot -= y;
    }
    
}
fn lca(mut a: usize, mut b: usize, tbl: &[Vec<usize>], dep: &[i32]) -> usize {
    if dep[a] > dep[b] {
        std::mem::swap(&mut a, &mut b);
    }
    if dep[a] < dep[b] {
        for i in (0..B).rev() {
            if dep[a] + (1 << i) <= dep[b] {
                b = tbl[i][b];
            }
        }
    }
    if a == b {
        return a;
    }
    for i in (0..B).rev() {
        if tbl[i][a] != tbl[i][b] {
            a = tbl[i][a];
            b = tbl[i][b];
        }
    }
    tbl[0][a]
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        n: usize, q: usize,
        abcd: [(usize1, usize1, usize, i64); n - 1],
        xyuv: [(usize, i64, usize1, usize1); q],
    }
    let mut g = vec![vec![]; n];
    let mut tbl = vec![vec![0; n]; B];
    let mut dep = vec![0; n];
    for &(a, b, c, d) in &abcd {
        g[a].push((b, c, d));
        g[b].push((a, c, d));
    }
    calc(0, n, &g, &mut tbl[0], &mut dep, 0);
    for i in 1..B {
        for j in 0..n {
            tbl[i][j] = tbl[i - 1][tbl[i - 1][j]];
        }
    }
    let mut ans = vec![0; q];
    let mut qs = vec![vec![]; n];
    for (i, &(x, y, u, v)) in xyuv.iter().enumerate() {
        let l = lca(u, v, &tbl, &dep);
        qs[l].push((x, y, -2, i));
        qs[u].push((x, y, 1, i));
        qs[v].push((x, y, 1, i));
    }
    let mut freq = vec![(0, 0); W];
    let mut tot = 0;
    dfs(0, n, &g, &qs, &mut ans,
        &mut freq, &mut tot);
    for i in 0..q {
        puts!("{}\n", ans[i]);
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}