#include <bits/stdc++.h>
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second

using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const int N = 3e5 + 50;
const ll mod = 998244353;
//__builtin_ctzll(x);后导0的个数
//__builtin_popcount计算二进制中1的个数
struct node {
    ll c, w, f;
};

bool cmp(node a, node b) {
    if (a.f == b.f) return a.w > b.w;
    return a.f > b.f;
}

void work() {
    ll n, k; cin >> n >> k;
    vector<node> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i].c >> a[i].w >> a[i].f;
    }
    sort(a.begin() + 1, a.end(), cmp);
    int r = 0;

    auto check = [&] -> ll {
        while (r <= n && (a[r].w != 1 || a[r].c <= 0)) r++;
        if (r <= n && a[r].w == 1 && a[r].c > 0) {
            a[r].c--; return 1;
        }
        return 0;
    };

    ll res = 0, ans = 0, nw = 0;
    for (int i = 1; i <= n; ++i) {
        //计算当前轮
        ll x = min((k - nw) / a[i].w, a[i].c);
        a[i].c -= x;
        nw += x * a[i].w;
        res = max(res, a[i].f);
        if (nw == k - 1) nw += check();
        if (nw >= k - 1) ans += res, res = 0, nw = 0;

        if (a[i].c == 0) continue;

        //加速后面的轮次
        x = a[i].c / (k / a[i].w);
        ans += x * a[i].f;
        a[i].c -= x * (k / a[i].w);

        if (a[i].c == 0) continue;

        //如果还有剩余
        nw = a[i].c * a[i].w;
        res = max(res, a[i].f);
        if (nw == k - 1) nw += check();
        if (nw >= k - 1) ans += res, res = 0, nw = 0;
    }
    if (res) ans += res;
    cout << ans << '\n';
}

signed main() {
    io;
    int t = 1;
    cin >> t;
    while (t--) {
        work();
    }
    return 0;
}