#include <bits/stdc++.h>
using namespace std;
// Speed up looking next line change with sparse table,
// and memoize based on [left_end, right_end] of each
// street. State is O(H + W), since for every line x and
// the section we enter it, and from that line we go left
// or right to change to another line y and z, then if we
// change the path and instead goes through a line that passes
// through y and z, we will never stop at x. Thus for each
// line only one "region" matters.
// Time: O(Q * (H + W))
using lint = long long;
const int MAXN = 100005;
int H, W, Q;
array<int, 3> A[MAXN];
int st[MAXN];
int et[MAXN];
lint dpl[MAXN];
lint dpr[MAXN];
int revpos[MAXN][2];
lint Solve(int X, int Y, int mask) {
lint res = 0;
int xl = 0, xr = H - 1;
int yl = 0, yr = W - 1;
memset(dpl, 0, sizeof(dpl));
memset(dpr, 0, sizeof(dpr));
for (int i = 0; i < H + W; i++) {
if (A[i][2] == 0) {
st[i] = yl, et[i] = yr;
if (revpos[st[i]][A[i][2] ^ 1] < i) {
int p = revpos[st[i]][A[i][2] ^ 1];
dpl[i] = max(dpl[p] + abs(st[p] - A[i][1]), dpr[p] + abs(et[p] - A[i][1]));
}
if (revpos[et[i]][A[i][2] ^ 1] < i) {
int p = revpos[et[i]][A[i][2] ^ 1];
dpr[i] = max(dpl[p] + abs(st[p] - A[i][1]), dpr[p] + abs(et[p] - A[i][1]));
}
if (A[i][1] + !!(mask & 1) <= X) {
xl = max(xl, A[i][1]);
} else {
xr = min(xr, A[i][1]);
}
} else if (A[i][2] == 1) {
st[i] = xl, et[i] = xr;
if (revpos[st[i]][A[i][2] ^ 1] < i) {
int p = revpos[st[i]][A[i][2] ^ 1];
dpl[i] = max(dpl[p] + abs(st[p] - A[i][1]), dpr[p] + abs(et[p] - A[i][1]));
}
if (revpos[et[i]][A[i][2] ^ 1] < i) {
int p = revpos[et[i]][A[i][2] ^ 1];
dpr[i] = max(dpl[p] + abs(st[p] - A[i][1]), dpr[p] + abs(et[p] - A[i][1]));
}
if (A[i][1] + !!(mask & 2) <= Y) {
yl = max(yl, A[i][1]);
} else {
yr = min(yr, A[i][1]);
}
}
if (A[i][1] == X && A[i][2] == 0) {
res = max(res, dpl[i] + abs(Y - st[i]));
res = max(res, dpr[i] + abs(Y - et[i]));
}
if (A[i][1] == Y && A[i][2] == 1) {
res = max(res, dpl[i] + abs(X - st[i]));
res = max(res, dpr[i] + abs(X - et[i]));
}
}
return res;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cin >> H >> W >> Q;
for (int i = 0; i < H; i++) {
cin >> A[i][0];
A[i][1] = i;
A[i][2] = 0;
}
for (int i = 0; i < W; i++) {
cin >> A[i + H][0];
A[i + H][1] = i;
A[i + H][2] = 1;
}
sort(begin(A), end(A));
reverse(begin(A), end(A));
for (int i = 0; i < H + W; i++) {
revpos[A[i][1]][A[i][2]] = i;
}
for (int i = 0; i < Q; i++) {
int S, T;
cin >> S >> T;
S--, T--;
cout << max({Solve(S, T, 0), Solve(S, T, 1),
Solve(S, T, 2), Solve(S, T, 3)}) << '\n';
}
return 0;
}
