#include <bits/stdc++.h>

using namespace std;

int depth[100005], father[100005], son[100005], size[100005], dis[100005];
struct Edge { int to, color, len; };
vector<Edge> edge[100005];

void dfs1(int x, int f)
{
    depth[x]  = depth[f] + 1;
    father[x] = f;
    son[x]    = 0;
    size[x]   = 1;
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == f) continue;
        dis[y] = dis[x] + edge[x][i].len;
        dfs1(y, x);
        size[x] += size[y];
        if (size[y] > size[son[x]]) son[x] = y;
    }
    return;
}

int peak[100005];

void dfs2(int x, int t)
{
    peak[x] = t;
    if (son[x] != 0) dfs2(son[x], t);
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == father[x] || y == son[x]) continue;
        dfs2(y, y);
    }
    return;
}

int lca(int x, int y)
{
    while (peak[x] != peak[y])
        if (depth[peak[x]] > depth[peak[y]])
            x = father[peak[x]];
        else y = father[peak[y]];
    return depth[x] < depth[y] ? x : y;
}

int ans[100005], color_sum[100005], color_cnt[100005];
struct Query { int idx, color, len, val; };
vector<Query> query[100005];

void dfs3(int x, int f)
{
    for (int i = 0, i_END_ = query[x].size(); i < i_END_; ++i) {
        int color = query[x][i].color, len = query[x][i].len;
        ans[query[x][i].idx] += (dis[x] - color_sum[color] +
                                 color_cnt[color] * len) * query[x][i].val;
    }
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == f) continue;
        ++color_cnt[edge[x][i].color];
        color_sum[edge[x][i].color] += edge[x][i].len;
        dfs3(y, x);
        --color_cnt[edge[x][i].color];
        color_sum[edge[x][i].color] -= edge[x][i].len;
    }
    return;
}

int n, q;

int main(void)
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i < n; ++i) {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        edge[a].push_back((Edge){b, c, d});
        edge[b].push_back((Edge){a, c, d});
    }
    dfs1(1, 0);
    dfs2(1, 1);
    for (int i = 1; i <= q; ++i) {
        int x, y, u, v;
        scanf("%d%d%d%d", &x, &y, &u, &v);
        query[u].push_back((Query){i, x, y, 1});
        query[v].push_back((Query){i, x, y, 1});
        query[lca(u, v)].push_back((Query){i, x, y, -2});
    }
    dfs3(1, 0);
    for (int i = 1; i <= q; ++i)
        printf("%d\n", ans[i]);
    return 0;
}
### 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 <bits/stdc++.h>

using namespace std;

int depth[100005], father[100005], son[100005], size[100005], dis[100005];
struct Edge { int to, color, len; };
vector<Edge> edge[100005];

void dfs1(int x, int f)
{
    depth[x]  = depth[f] + 1;
    father[x] = f;
    son[x]    = 0;
    size[x]   = 1;
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == f) continue;
        dis[y] = dis[x] + edge[x][i].len;
        dfs1(y, x);
        size[x] += size[y];
        if (size[y] > size[son[x]]) son[x] = y;
    }
    return;
}

int peak[100005];

void dfs2(int x, int t)
{
    peak[x] = t;
    if (son[x] != 0) dfs2(son[x], t);
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == father[x] || y == son[x]) continue;
        dfs2(y, y);
    }
    return;
}

int lca(int x, int y)
{
    while (peak[x] != peak[y])
        if (depth[peak[x]] > depth[peak[y]])
            x = father[peak[x]];
        else y = father[peak[y]];
    return depth[x] < depth[y] ? x : y;
}

int ans[100005], color_sum[100005], color_cnt[100005];
struct Query { int idx, color, len, val; };
vector<Query> query[100005];

void dfs3(int x, int f)
{
    for (int i = 0, i_END_ = query[x].size(); i < i_END_; ++i) {
        int color = query[x][i].color, len = query[x][i].len;
        ans[query[x][i].idx] += (dis[x] - color_sum[color] +
                                 color_cnt[color] * len) * query[x][i].val;
    }
    for (int i = 0, i_END_ = edge[x].size(); i < i_END_; ++i) {
        int y = edge[x][i].to;
        if (y == f) continue;
        ++color_cnt[edge[x][i].color];
        color_sum[edge[x][i].color] += edge[x][i].len;
        dfs3(y, x);
        --color_cnt[edge[x][i].color];
        color_sum[edge[x][i].color] -= edge[x][i].len;
    }
    return;
}

int n, q;

int main(void)
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i < n; ++i) {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        edge[a].push_back((Edge){b, c, d});
        edge[b].push_back((Edge){a, c, d});
    }
    dfs1(1, 0);
    dfs2(1, 1);
    for (int i = 1; i <= q; ++i) {
        int x, y, u, v;
        scanf("%d%d%d%d", &x, &y, &u, &v);
        query[u].push_back((Query){i, x, y, 1});
        query[v].push_back((Query){i, x, y, 1});
        query[lca(u, v)].push_back((Query){i, x, y, -2});
    }
    dfs3(1, 0);
    for (int i = 1; i <= q; ++i)
        printf("%d\n", ans[i]);
    return 0;
}
```