#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define forr(i,a) for(auto i:a)
#define all(a) begin(a),end(a)
#define rall(a) rbegin(a),rend(a)
#define pb emplace_back
using namespace std;
struct read
{
    static const int M = 1 << 23;
    char buf[M], *S = buf, *P = buf, c, l;
    inline char gc() { return (S == P && (P = (S = buf) + fread(buf, 1, M, stdin), S == P) ? EOF : *S++); }
    template<class T> inline read &operator>>(T &x)
    {
        for (c = 0;!isdigit(c);c = gc()) l = c;
        for (x = 0;isdigit(c);c = gc()) x = x * 10 + (c & 15);
        return x = (l ^ 45) ? x : -x, *this;
    }
}Cin;
struct T
{
    int x, y, w;
    bool operator<(T &_) { return w < _.w; }
};
constexpr int N = 1e5 + 10;
T a[N]; int mx[N][4];
void solve()
{
    int n, q; Cin >> n >> q;
    rep(i, 1, n)
    {
        auto &[x, y, w] = a[i];
        Cin >> x >> y >> w;
    }
    sort(a + 1, a + n + 1);
    memset(mx[n + 1], -0x3f, sizeof(mx[n + 1]));
    per(i, n, 1)
    {
        auto &[x, y, w] = a[i];
        mx[i][0] = max(mx[i + 1][0], x + y);
        mx[i][1] = max(mx[i + 1][1], x - y);
        mx[i][2] = max(mx[i + 1][2], -x + y);
        mx[i][3] = max(mx[i + 1][3], -x - y);
    }
    for (int x, y;q--;)
    {
        Cin >> x >> y;
        int l = 1, r = n, ret = INT_MIN;
        while (l <= r)
        {
            int mid = l + r >> 1,
                cur = max({ x + y + mx[mid][3], x - y + mx[mid][2], -x + y + mx[mid][1], -x - y + mx[mid][0] });
            auto &[X, Y, W] = a[mid];
            if (cur <= W) r = mid - 1, ret = max(ret, cur);
            else l = mid + 1, ret = max(ret, W);
        }
        printf("%d\n", ret);
    }
}
signed main()
{
    int _;for (Cin >> _; _--;) solve();
    return 0;
}