// Original C++ solution (slower):
/*

How to lambda

  ex) sort(ALL(cont),[](dat a,dat b){return a.l<b.l;});

lower_bound/upper_bound

  1 2 2 2 3 3 

    ^lb   ^ub */

#include <bits/stdc++.h>

#define BR "\n"

#define REP(i,n) for(int (i)=0;(i)<(n);++(i))

#define ALL(cont) begin(cont),end(cont)

#define AS_MOD(a,b) ((((a) % (b) ) + (b)) % (b))

#define FEACH(it,cont) for(auto (it) = begin(cont); it!=end(cont);++it)

#define FEACHR(it,cont) for(auto (it) = rbegin(cont); it!=rend(cont);++it)

#define pb push_back

#define pob pop_back

#define fi first

#define se second

#define getll() ([](){ll s;scanf("%lld", &s);return s;})()

#define getld() ([](){ld s;scanf("%Lf", &s);return s;})()

#define prl(P) printf("%lld", P)

#define prd(P) printf("%.10Lf", P)

#define pr(P) printf(P)

#define assign_if_greater(V, T) V=max(V,T)

using namespace std;

typedef long long ll;

typedef long double ld;

int A,B,Q;

ll As[100000];

ll Bs[100000];

ll INF=(ll)1000*1000*1000*1000;

int main() {



  scanf("%d%d%d", &A, &B, &Q);

  REP(i,A)

    As[i]=getll();

  REP(i,B)

    Bs[i]=getll();

  REP(i,Q) {

    ll x=getll();

    ll* ita=lower_bound(As, As+A, x);

    ll* itb=lower_bound(Bs, Bs+B, x);

    ll ans=INF;

    vector<ll> alis,blis;

    if (ita!=As+A) {

      alis.pb(*ita);

    }

    if (itb!=Bs+B) {

      blis.pb(*itb);

    }

    if (ita!=As) {

      ita--;

      alis.pb(*ita);

    }

    if (itb!=Bs) {

      itb--;

      blis.pb(*itb);

    }

    for(ll a:alis)

      for(ll b:blis) {

        ans = min(ans,abs(a-x)+abs(b-a));

        ans = min(ans,abs(b-x)+abs(b-a));

      }

    printf("%lld\n",ans);

  }





  return 0;

}


// Optimized solution to the same problem (faster):
/*

How to lambda

  ex) sort(ALL(cont),[](dat a,dat b){return a.l<b.l;});

lower_bound/upper_bound

  1 2 2 2 3 3 

    ^lb   ^ub */

#include <algorithm>

#include <cstdio>

#include <cstdlib>

#define BR "\n"

#define REP(i,n) for(int (i)=0;(i)<(n);++(i))

#define ALL(cont) begin(cont),end(cont)

#define AS_MOD(a,b) ((((a) % (b) ) + (b)) % (b))

#define FEACH(it,cont) for(auto (it) = begin(cont); it!=end(cont);++it)

#define FEACHR(it,cont) for(auto (it) = rbegin(cont); it!=rend(cont);++it)

#define pb push_back

#define pob pop_back

#define fi first

#define se second

#define getll() ([](){ll s;scanf("%lld", &s);return s;})()

#define getld() ([](){ld s;scanf("%Lf", &s);return s;})()

#define prl(P) printf("%lld", P)

#define prd(P) printf("%.10Lf", P)

#define pr(P) printf(P)

#define assign_if_greater(V, T) V=max(V,T)

using namespace std;

typedef long long ll;

typedef long double ld;

int A,B,Q;

ll As[100000];

ll Bs[100000];

ll INF=(ll)1000*1000*1000*1000;

ll fastGet() {

  ll ret=0;

  char c=getchar_unlocked();

  // skipping

  while(c<'0' || c > '9') c=getchar_unlocked();

  do {

    ret *= 10;

    ret += (c-'0');

    c = getchar_unlocked();

  } while(c>='0' && c<='9');

  return ret;

}

void fastPut(ll v) {

  char s[256];

  int p=0;

  do {

    s[p++]=(v%10)+'0';

    v/=10;

  } while(v>0);

  while(p--)

    putchar_unlocked(s[p]);

}

int main() {

  A=fastGet();

  B=fastGet();

  Q=fastGet();

  REP(i,A)

    As[i]=fastGet();

  REP(i,B)

    Bs[i]=fastGet();

  REP(i,Q) {

    ll x=fastGet();

    ll* ita=lower_bound(As, As+A, x);

    ll* itb=lower_bound(Bs, Bs+B, x);

    ll ans=INF;

    ll alis[2],blis[2];

    alis[0]=ita!=As+A?*ita:INF;

    blis[0]=itb!=Bs+B?*itb:INF;

    alis[1]=ita!=As?ita[-1]:-INF;

    blis[1]=itb!=Bs?itb[-1]:-INF;

    for(ll a:alis)

      for(ll b:blis) {

        ans = min(ans,abs(a-x)+abs(b-a));

        ans = min(ans,abs(b-x)+abs(b-a));

      }

    fastPut(ans);

    putchar_unlocked('\n');

  }

  return 0;

}


// Fastest known solution for this problem: