样例全过WA0pts

#include <bits/stdc++.h>
using namespace std;

//For the basic const define.
const int MAXN = 1e5 + 10;
const int LOGN = 21;
const int inf =  2147483647;
const int inm = -2147483647;

//For the basic define.
int n, m, q;
int a[MAXN], b[MAXN];

//For the maximum of a and b.
int mxsta[MAXN][LOGN], mxstb[MAXN][LOGN];

//For the minimum of a and b.
int mnsta[MAXN][LOGN], mnstb[MAXN][LOGN];

//For the minimum which bigger then 0 of a.
int mnbsta[MAXN][LOGN];

//For the maximum which smaller then 0 of a.
int mxssta[MAXN][LOGN];

//For the quick log2.
int logx[MAXN];

//For the quick read of inteeger.
inline int read() {
	char ch = getchar();
	int x = 0, f = 1;
	while (ch < '0' || ch > '9') {
		if (ch == '-')
			f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar();
	}
	return x * f;
}

//For the initialization.
inline void init() {
	//Read.
	n = read();
	m = read();
	q = read();
	for (int i = 1; i <= n; i++) {
		mnsta[i][0] = mxsta[i][0] = a[i] = read();
		if (a[i] >= 0) {
			mnbsta[i][0] = a[i];
			mxssta[i][0] = inm;
		} else {
			mxssta[i][0] = a[i];
			mnbsta[i][0] = inf;
		}
	}
	for (int i = 1; i <= m; i++)
		mnstb[i][0] = mxstb[i][0] = b[i] = read();

	//Prepare for the quick log2.
	logx[1] = 0;
	logx[2] = 1;
	for (int i = 3; i < MAXN; i++)
		logx[i] = logx[i / 2] + 1;

	//Prepare for Sprase Tables.
	for (int j = 1; j <= LOGN; j++) {
		for (int i = 1; i + (1 << j) - 1 <= n; i++) {
			int s = i + (1 << (j - 1));
			mxsta[i][j] = max(mxsta[i][j - 1], mxsta[s][j - 1]);
			mxssta[i][j] = max(mxssta[i][j - 1], mxssta[s][j - 1]);
			mnsta[i][j] = min(mnsta[i][j - 1], mnsta[s][j - 1]);
			mnbsta[i][j] = min(mnbsta[i][j - 1], mnbsta[s][j - 1]);
		}
		for (int i = 1; i + (1 << j) - 1 <= m; i++) {
			int s = i + (1 << (j - 1));
			mxstb[i][j] = max(mxstb[i][j - 1], mxstb[s][j - 1]);
			mnstb[i][j] = min(mnstb[i][j - 1], mnstb[s][j - 1]);
		}
	}
}

//For the O(1) query of max.
inline int max_query(int l, int r, int f[MAXN][LOGN]) {
	int s = logx[r - l + 1];
	return max(f[l][s], f[r - (1 << s) + 1][s]);
}

//For the O(1) query of min.
inline int min_query(int l, int r, int f[MAXN][LOGN]) {
	int s = logx[r - l + 1];
	return min(f[l][s], f[r - (1 << s) + 1][s]);
}

//Update the best idea.
void update(int &a, int b) {
	if (b > a)
		a = b;
}

//Solve the problem.
inline void solve() {
	int l1, r1, l2, r2;
	l1 = read(), r1 = read();
	l2 = read(), r2 = read();
	int maxa = max_query(l1, r1, mxsta);
	int mina = min_query(l1, r1, mnsta);
	int mxsa = max_query(l1, r1, mxssta);
	int mnba = min_query(l1, r1, mnbsta);
	int maxb = max_query(l2, r2, mxstb);
	int minb = min_query(l2, r2, mnstb);
	int ans = inm;
	update(ans, maxa * (maxa >= 0 ? minb : maxb));
	update(ans, mina * (mina >= 0 ? minb : maxb));
	if (mxsa != inm)
		update(ans, mxsa * (mxsa >= 0 ? minb : maxb));
	if (mnba != inf)
		update(ans, mnba * (mnba >= 0 ? minb : maxb));
	printf("%d\n", ans);
}

int main() {
	//Initialization.
	init();

	//For q times use the solve function.
	while (q--)
		solve();
	return 0;
}