#include <iostream>
#include <queue>
#include <cmath>
using namespace std;

//最近公共祖先问题 lca问题
const int N = 1e5 + 10, M = 2 * N;
//先存树
int h[N], e[M], ne[M], w[M], idx;
int T, n, m, t;

void add(int a, int b, int c)
{
    e[++idx] = b;
    ne[idx] = h[a];
    h[a] = idx;
}

//预处理
int f[N][20], d[N], dist[N];
queue<int> q;
void bfs()//bfs求f和d数组
{
    q.push(1);
    d[1] = 1;

    while (q.size())
    {
        int x = q.front();
        q.pop();

        for (int i = h[x]; i != -1; i = ne[i])
        {
            int y = e[i];
            if (d[y])
                continue;
            d[y] = d[x] + 1;
            dist[y] = dist[x] + w[i];

            f[y][0] = x;

            for (int j = 1; j <= t; j++)
                f[y][j] = f[f[y][j - 1]][j - 1];

            q.push(y);
        }
    }
}

//dfs求f和d数组 int f[N][20], d[N], dist[N];
void dfs(int u, int father, int ver) //u为当前遍历到的结点,father为他的父节点
{
    f[u][0] = father;
    dist[u] = dist[father] + ver; //ver为father到u的边
    d[u] = d[father] + 1;
    //迭代求祖先
    for (int i = 1; i <= t; i++)
        f[u][i] = f[f[u][i - 1]][i - 1];

    //dfs求儿子
    for (int i = u; i != -1; i = ne[i])
    {
        int j = e[i];
        if (j != father)
            dfs(j, u, w[i]);
    }
}

long long lca(int x, int y)
{
    if (d[x] > d[y])
        swap(x, y);
    for (int i = t; i >= 0; i--)
    {
        if (d[f[y][i]] >= d[x])
            y = f[y][i];
    }
    if (x == y)
        return x;
    for (int i = t; i >= 0; i--)
        if (f[x][i] != f[y][i])
            x = f[x][i], y = f[y][i];

    return f[x][0];
}


int main()
{
    ios::sync_with_stdio(false);
    cin >> T;
    while (T--)
    {
        cin >> n >> m;
        t = (int) (log(n) / log(2)) + 1;//一般小于19

        for (int i = 1; i <= n; i++)
            h[i] = d[i] = 0;
        idx = 0;

        for (int i = 1; i < n; i++)
        {
            int x, y, z;
            cin >> x >> y >> z;
            add(x, y, z);
            add(y, x, z);
        }

        bfs();

        for (int i = 1; i <= m; i++)
        {
            int x, y;
            cin >> x >> y;

            printf("%d\n", dist[x] + dist[y] - 2 * dist[lca(x, y)]);
        }
    }

    return 0;
}