#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#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 ls (u<<1)
#define rs (u<<1|1)
#define mid ((l+r)>>1)
using namespace std;
typedef long long ll;
const int N = 1000010;

class TREE{
	struct node{
		ll mn, lzy;
		void tag(ll x) {mn += x, lzy += x;}
	};
	node t[N<<2];
	void up(int u) {t[u].mn = min(t[ls].mn, t[rs].mn);}
	void down(int u){
		if(!t[u].lzy) return;
		t[ls].tag(t[u].lzy), t[rs].tag(t[u].lzy), t[u].lzy = 0;
	}
	public:
	void build(int u, int l, int r, ll *a){
		t[u].lzy = 0;
		if(l == r) t[u].mn = a[l];
		else build(ls, l, mid, a), build(rs, mid + 1, r, a), up(u);
	}
	void update(int u, int l, int r, int x, int y, ll val){
		if(x <= l && r <= y) return t[u].tag(val);
		down(u);
		if(x <= mid) update(ls, l, mid, x, y, val);
		if(y > mid) update(rs, mid + 1, r, x, y, val);
		up(u);
	}
	ll query(int u, int l, int r, int x, int y){
		if(x <= l && r <= y) return t[u].mn;
		down(u);
		if(y <= mid) return query(ls, l, mid, x, y);
		if(x > mid) return query(rs, mid + 1, r, x, y);
		return min(query(ls, l, mid, x, y), query(rs, mid + 1, r, x, y));
	}
};
TREE tr;

int n;
ll m, c, a[N], b[N], ans, s[N], x[N];

void solve1(){
	s[1] = m;
	rep(i, 1, n){
		ll t = min({s[i], a[i] % c, b[i]});
		ans -= t, x[i] += t;
		s[i+1] = s[i] - t + (a[i] - t) / c;
	}
}

void solve2(){
	ll mn = s[n+1];
	per(i, n, 1){
		ll t = min({(b[i] - x[i]) / c, (s[i] - x[i]) / c, mn / (c+1)});
		x[i] += t * c, ans -= t * c;
		mn = min(mn - t * (c+1), s[i] - x[i]);
	}
	// printf("solve2: %lld\n", mn);
	s[1] = m;
	rep(i, 1, n) s[i+1] = s[i] - x[i] + (a[i] - x[i]) / c;
}

priority_queue<int> q;

void solve3(){
	static int id[N];
	static ll tmp[N], rk[N];
	rep(i, 1, n+1) tmp[i] = s[i] - x[i];
	tr.build(1, 1, n+1, tmp);
	rep(i, 1, n) id[i] = i, rk[i] = min(c-1, b[i] - x[i]);
	id[n+1] = n+1, rk[n+1] = 0;
	sort(id+1, id+n+1, [&](int x, int y) {return rk[x] > rk[y];});
	for(int l = 1, r = 0; l <= n; l = r + 1){
		while(r < n && rk[id[r+1]] == rk[id[l]]) q.push(id[++r]);
		while(!q.empty()){
			int x = q.top();
			ll t = min(tr.query(1, 1, n+1, x, x), tr.query(1, 1, n+1, x+1, n+1) - 1);
			if(t <= rk[id[r+1]]) break;
			t = min(t, c-1);
			ans -= t; q.pop();
			tr.update(1, 1, n+1, x, x, -t);
			tr.update(1, 1, n+1, x+1, n+1, -t-1);
		}
	}
}

void solve(){
	while(!q.empty()) q.pop();
	ans = 0;
	rep(i, 1, n) x[i] = 0;
	scanf("%d%lld%lld", &n, &m, &c);
	rep(i, 1, n) scanf("%lld", &a[i]), ans += a[i];
	rep(i, 1, n) scanf("%lld", &b[i]);
	solve1();
	solve2();
	solve3();
	printf("%lld\n", ans);
}

int main(){
	int T; scanf("%d", &T); while(T--) solve();
	return 0;
}