Problem:
Score : 400 points

Problem Statement

Along a road running in an east-west direction, there are A shrines and B temples.
The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.

Answer the following Q queries:

Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)

Constraints

1 \leq A, B \leq 10^5

1 \leq Q \leq 10^5

1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}

1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}

1 \leq x_i \leq 10^{10}

s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

2 3 4
100
600
400
900
1000
150
2000
899
799

Sample Output 1

350
1400
301
399

There are two shrines and three temples. The shrines are located at distances of 100, 600 meters from the west end of the road, and the temples are located at distances of 400, 900, 1000 meters from the west end of the road.

Query 1: If we start from a point at a distance of 150 meters from the west end of the road, the optimal move is first to walk 50 meters west to visit a shrine, then to walk 300 meters east to visit a temple.

Query 2: If we start from a point at a distance of 2000 meters from the west end of the road, the optimal move is first to walk 1000 meters west to visit a temple, then to walk 400 meters west to visit a shrine. We will pass by another temple on the way, but it is fine.

Query 3: If we start from a point at a distance of 899 meters from the west end of the road, the optimal move is first to walk 1 meter east to visit a temple, then to walk 300 meters west to visit a shrine.

Query 4: If we start from a point at a distance of 799 meters from the west end of the road, the optimal move is first to walk 199 meters west to visit a shrine, then to walk 200 meters west to visit a temple.

Sample Input 2

1 1 3
1
10000000000
2
9999999999
5000000000

Sample Output 2

10000000000
10000000000
14999999998

The road is quite long, and we may need to travel a distance that does not fit into a 32-bit integer.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#if 0
cat <<EOF >mistaken-paste
#endif
// thanks for @rsk0315_h4x

#pragma GCC diagnostic ignored "-Wincompatible-pointer-types"
#define _USE_MATH_DEFINES

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <time.h>

#define BIG 2000000007
#define VERYBIG 2000000000000007LL

#define MOD 1000000007
#define FOD  998244353
typedef uint64_t ull;
typedef  int64_t sll;

#define N_MAX 1048576

#ifdef __cplusplus
#include <queue>
#include <stack>
#include <tuple>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <functional>
#include <array>

using std::queue;
using std::priority_queue;
using std::stack;
using std::tuple;
using std::set;
using std::map;
using std::vector;
using std::greater;
using std::pair;
using std::string;
using std::get;

template <typename T, typename U>
pair<T, U> operator+(pair<T, U> l, pair<T, U> r) {
	return pair<T, U>(l.first + r.first, l.second + r.second);
}

#endif

typedef struct {
	int32_t a;
	int32_t b;
} hw;

typedef struct {
	sll a;
	sll b;
} hwll;

typedef struct {
	sll a;
	sll b;
	sll c;
} hwllc;

typedef struct {
	hwll a;
	hwll b;
} linell;

ull n, m;
ull h, w;
ull k;
ull q;
sll va, vb, vc, vd, ve, vf;
ull ua, ub, uc, ud, ue, uf;
long double vra, vrb, vrc;
double vda, vdb, vdc;
char ch, dh;

ull umin (ull x, ull y) {
	return (x < y) ? x : y;
}

ull umax (ull x, ull y) {
	return (x > y) ? x : y;
}

sll smin (sll x, sll y) {
	return (x < y) ? x : y;
}

sll smax (sll x, sll y) {
	return (x > y) ? x : y;
}

ull gcd (ull x, ull y) {
	if (x < y) {
		return gcd(y, x);
	} else if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}

ull bitpow (ull a, ull x, ull modulo) {
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= modulo;
		}
		x /= 2;
		a = (a * a) % modulo;
	}
	return result;
}

ull divide (ull a, ull b, ull modulo) {
	return (a * bitpow(b, modulo - 2, modulo)) % modulo;
}

ull udiff (ull a, ull b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

sll sdiff (sll a, sll b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

int bitcount (ull n) {
	int result = 0;
	while (n) {
		if (n & 1) result++;
		n /= 2;
	}
	return result;
}

#define BEGCMP(NAME) int32_t NAME (const void *left, const void *right)
#define DEFLR(TYPE) TYPE l=*(TYPE*)left,r=*(TYPE*)right
#define CMPRET(L, R) if((L)<(R))return-1;if((L)>(R))return+1

// int32_t pullcomp (const void *left, const void *right) {
// 	ull l = *(ull*)left;
// 	ull r = *(ull*)right;
// 	if (l < r) {
// 		return -1;
// 	}
// 	if (l > r) {
// 		return +1;
// 	}
// 	return 0;
// }
BEGCMP(pullcomp){
	DEFLR(ull);
	CMPRET(l, r);
	return 0;
}
BEGCMP(prevcomp){
	DEFLR(ull);
	CMPRET(r, l);
	return 0;
}
BEGCMP(psllcomp){
	DEFLR(sll);
	CMPRET(l, r);
	return 0;
}
BEGCMP(pcharcomp){
	DEFLR(char);
	CMPRET(l, r);
	return 0;
}
BEGCMP(pdoublecomp){
	DEFLR(double);
	CMPRET(l, r);
	return 0;
}

int32_t pstrcomp (const void *left, const void *right) {
	char* l = *(char**)left;
	char* r = *(char**)right;

	return strcmp(l, r);
}

BEGCMP(phwllABcomp){
	DEFLR(hwll);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	return 0;
}
BEGCMP(phwllREVcomp){
	DEFLR(hwll);
	CMPRET(l.b, r.b);
	CMPRET(l.a, r.a);
	return 0;
}
BEGCMP(ptriplecomp){
	DEFLR(hwllc);
	CMPRET(l.a, r.a);
	CMPRET(l.b, r.b);
	CMPRET(l.c, r.c);
	return 0;
}
BEGCMP(ptripleREVcomp){
	DEFLR(hwllc);
	CMPRET(l.b, r.b);
	CMPRET(l.a, r.a);
	CMPRET(l.c, r.c);
	return 0;
}

int32_t pquadcomp (const void *left, const void *right) {
	linell l = *(linell*)left;
	linell r = *(linell*)right;

	sll ac = phwllABcomp(&(l.a), &(r.a));
	if (ac) return ac;
	sll bc = phwllABcomp(&(l.b), &(r.b));
	if (bc) return bc;

	return 0;
}

bool isinrange (sll left, sll x, sll right) {
	return (left <= x && x <= right);
}

bool isinrange_soft (sll left, sll x, sll right) {
	return (left <= x && x <= right) || (left >= x && x >= right);
}

sll a[N_MAX + 5];
// ull a[N_MAX + 5];
// sll a[3001][3001];
sll b[N_MAX + 5];
// ull b[N_MAX + 5];
// sll b[3001][3001];
sll c[N_MAX + 5];
sll d[N_MAX + 5];
sll e[N_MAX];
char s[N_MAX + 1];
// char s[3010][3010];
char t[N_MAX + 1];
// char t[3010][3010];
char u[N_MAX + 1];
hwll xy[N_MAX + 5];
hwllc tup[N_MAX + 5];
// sll table[3005][3005];
ull gin[N_MAX];
// here we go

bool isok (ull x) {
	ull y = 0;
	while (x > 2) {
		x = (x - 1) / 2;
		y++;
	}
	return (y <= n);
}

ull solve () {
	sll i, j, ki, li;
	// ull result = 0;
	sll result = 0;
	double dresult = 0;
	// ull maybe = 0;
	sll maybe = 0;
	// ull sum = 0;
	sll sum = 0;
	sll item;
	ull *dpcell;

	for (i = n; i; i--) {
		a[i] = a[i - 1];
	}
	a[0] = -VERYBIG;
	a[n + 1] = +VERYBIG;
	for (i = m; i; i--) {
		b[i] = b[i - 1];
	}
	b[0] = -VERYBIG;
	b[m + 1] = +VERYBIG;

	while (q--) {
		sll x;
		scanf("%lld", &x);
		ull ok1 = 0, ng1 = n + 1;
		while (ok1 + 1 < ng1) {
			ull med = (ok1 + ng1) / 2;
			if (a[med] <= x) {
				ok1 = med;
			} else {
				ng1 = med;
			}
		}
		ull ok2 = 0, ng2 = m + 1;
		while (ok2 + 1 < ng2) {
			ull med = (ok2 + ng2) / 2;
			if (b[med] <= x) {
				ok2 = med;
			} else {
				ng2 = med;
			}
		}

		sll l1 = smin(x - a[ok1], x - b[ok2]), l2 = smax(x - a[ok1], x - b[ok2]);
		sll r1 = smin(a[ng1] - x, b[ng2] - x), r2 = smax(a[ng1] - x, b[ng2] - x);
		result = smin(smin(l2, r2), l1 + r1 + smin(l1, r1));
		if (result < 0) abort();
		printf("%lld\n", result);
	}

	// printf("%lld\n", result);
	// printf("%.15lf\n", dresult);
	// puts(s);

	return 0;

	success:
	// puts("YES");
	puts("Yes");
	// printf("%llu\n", result);
	// puts("0");
	// puts("Hanako");
	return 0;

	fail:
	// puts("NO");
	// puts("No");
	puts("0");
	// puts("-1");
	// puts("-1 -1 -1");
	// puts("Jiro");
	return 1;
}

int32_t main (int argc, char *argv[]) {
	int32_t i, j;

	n = 3;
	m = 0;

	// scanf("%llu%llu", &h, &w);
	scanf("%llu%llu", &n, &m);
	// scanf("%llu", &k, &n, &m);
	// scanf("%llu%llu", &h, &w);
	// scanf("%s", s);
	scanf("%llu", &q);
	// scanf("%lld%lld", &va, &vb, &vc, &vd);
	// scanf("%llu%llu%llu", &ua, &ub, &uc, &ud, &ue);
	// scanf("%s", t);
	// scanf("%s", u);
	// scanf("%llu", &k);
	// scanf("%lld", &m);
	// for (i = 0; i < n; i++) {
	// 	scanf("%llu", &d[i]);
	// }
	for (i = 0; i < n; i++) {
		// scanf("%lld%lld", &xy[i].a, &xy[i].b);
		// scanf("%lld%lld%lld", &tup[i].a, &tup[i].b, &tup[i].c);
		scanf("%lld", &a[i]);
		// scanf("%lld", &b[i]);
		// scanf("%lld", &c[i]);
		// scanf("%lld", &d[i]);
		// a[i]--;
		// b[i]--;
		// c[i]--;
		// d[i]--;
		// xy[i].a--;
		// xy[i].b--;
		// tup[i].a--;
		// tup[i].b--;
	}
	for (i = 0; i < m; i++) {
		// scanf("%lld%lld", &a[i], &b[i]);
		scanf("%lld", &b[i]);
		// a[i]--;
		// b[i]--;
	}

	// for (i = 0; i < q; i++) {
	// 	// scanf("%lld%lld", &xy[i].a, &xy[i].b);
	// 	scanf("%lld", &c[i]);
	// 	// xy[i].a--;
	// 	// xy[i].b--;
	// }

	// for (i = 0; i < n; i++) {
	// 	for (j = 0; j < m; j++) {
	// 		scanf("%lld", &table[i][j]);
	// 		// table[i][j]--;
	// 	}
	// }
	// for (i = 0; i < h; i++) {
	// 	scanf("%s", s[i]);
	// }
	// scanf("%llu", &q);

	solve();

	return 0;
}