//蒟蒻一枚 rp++
//即得易见平凡，仿照上例显然。留作习题答案略，读者自证不难
//反之亦然同理，推论自然成立，略去过程Q.E.D.，由上可知证毕
#include<bits/stdc++.h>
//#pragma GCC optimize("Ofast")
#define re register
#define il inline
#define gc() getchar()
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define repp(i,a,b) for(int i=(a);i<(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define tep(i,x) for(int i=head[x];~i;i=ne[i])
#define ls(x) x<<1
#define rs(x) x<<1|1
#define epos (1e-9)
#define inf 0x3f3f3f3f
#define INF 1e16
#define pii pair<int,int>
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define fi first
#define sc second
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef double db;
namespace IO{
// #define gc()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<9,stdin),p1==p2)?EOF:*p1++)
// char buf[1<<9],*p1=buf,*p2=buf;
template<typename T> inline void read(T &x){
bool f=1;x=0;char ch=gc();
while(ch<'0'||ch>'9'){if(ch=='-')f=0;ch=gc();}
while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+(ch&15),ch=gc();
x=f?x:-x;
}
template<typename T> inline void write(T x){
if(x<0) putchar('-'),x=-x;
if(x>9) write(x/10);
putchar(char('0'+x%10));
}
template <typename T,typename ...Args> inline
void read(T &x,Args &...args){read(x);read(args...);}
template<typename T> inline void write(T x,char c){write(x),putchar(c);}
}
using namespace IO;
namespace MOD{
typedef uint32_t u32;
typedef uint64_t u64;
typedef __uint128_t u128;
u64 p,m;
void set_mod(u32 mod){
m=mod,p=-m/m+1;
}
u64 mo(u64 x){//取模优化
x-=u64((u128)x*p>>64)*m;
return (x>=m)?x-m:x;
}
}
bool _ST;
#define int LL
const int N=1e5+100;
struct NODE{
int len,pos,cnt1,cnt2,ori,type;
NODE(int _len=0,int _pos=0,int _cnt1=0,int _cnt2=0,int _ori=0,int _type=0){
len=_len,pos=_pos,cnt1=_cnt1,cnt2=_cnt2,ori=_ori,type=_type;
}
bool operator <(const NODE&T)const{
return (len==T.len)?pos>T.pos:len<T.len;
}
};
int n,q;
LL m,pos[N],a[N],ans[N];
priority_queue<NODE> que;
pii qu[N];
LL calc1(LL len,LL k,LL t){
LL res=0;
while(len>1){
if(t<=(k>>1)) len>>=1,k>>=1;
else res+=len>>1,len>>=1,t-=(k>>1),k=(k+1)>>1;
}
return res+1;
}
LL calc2(LL len,LL k,LL t){
LL res=0;
while(len>1){
if(t<=len/2-(k>>1)) len>>=1,k>>=1;
else res+=len>>1,t-=len/2-(k>>1),len>>=1,k=(k+1)>>1;
}
return res+1;
}
LL calc3(LL len,LL k,LL t){
LL res=0;
while(len>1){
if(t<=len/2) len>>=1,k>>=1;
else res+=k>>1,t-=len/2,len>>=1,k=(k+1)>>1;
}
return res+k;
}
bool _ED;
signed main(){
fprintf(stderr,"%.20lf MB\n",(&_ST-&_ED)/1048576.0);
//ios::sync_with_stdio(false);
//cin.tie(0);cout.tie(0);
read(m,n,q);
rep(i,1,n) read(a[i]),pos[i]=a[i];
sort(a+1,a+1+n);a[n+1]=m+1;
rep(i,1,n+1) if(a[i]-a[i-1]-1>0) que.push(NODE(a[i]-a[i-1]-1,a[i-1]+1,1,0,1,0));
rep(i,1,q) read(qu[i].fi),qu[i].sc=i;
sort(qu+1,qu+q+1);
int cur=1;
while(cur<=q&&qu[cur].fi<=n) ans[qu[cur].sc]=pos[qu[cur].fi],++cur;
LL ti=n;
while(!que.empty()&&cur<=q){
auto u=que.top();que.pop();
if(!u.len) break;
LL ed=ti+u.cnt1;
//cerr<<ti<<" "<<ed<<endl;
if(u.type==2) ed+=u.cnt2;
while(cur<=q&&qu[cur].fi<=ed){
if(u.type==0){
LL res=calc1(u.ori,u.cnt1,qu[cur].fi-ti);
ans[qu[cur].sc]=u.pos+(qu[cur].fi-ti-1)*(u.len+1)+(res-(qu[cur].fi-ti))*u.len+(u.len-1)/2;
}
else if(u.type==1){
LL res=calc2(u.ori,u.cnt2,qu[cur].fi-ti);
ans[qu[cur].sc]=u.pos+(qu[cur].fi-ti-1)*(u.len+1)+(res-(qu[cur].fi-ti))*(u.len+2)+(u.len-1)/2;
}
else{//1
LL res=calc3(u.ori,u.cnt1,qu[cur].fi-ti);
ans[qu[cur].sc]=u.pos+2ll*(qu[cur].fi-ti-1)+res;
}
++cur;
}
ti=ed;
if(u.type==0){
if(u.len==2){
que.push(NODE(1,u.pos,u.cnt1,u.cnt2,u.ori,2));
}
else{
que.push(NODE(u.len-1,u.pos,u.cnt2,u.cnt1,u.ori,1));
}
}
if(u.type==1){
int cnt1,cnt2;
if(u.len%2==0) cnt1=u.cnt1+u.cnt2*2,cnt2=u.cnt1;//0->00,1->01
else cnt1=u.cnt2,cnt2=u.cnt1*2+u.cnt2;
que.push(NODE((u.len+1)/2,u.pos,cnt1,cnt2,u.ori*2ll,0));
}
}
rep(i,1,q) write(ans[i],'\n');
fprintf(stderr,"%.4lf s\n",1.0*clock()/CLOCKS_PER_SEC);
return 0;
}
// ./seat < seat3.in > seat.out
//谨记:
//十年OI一场空，不开longlong见祖宗
//数据千万条，清空第一条。多测不清空，爆零两行泪。清空不规范，TLE总相伴。
//快读要加类型名
