We have a sandglass consisting of two bulbs, bulb A and bulb B. These bulbs contain some amount of sand. When we put the sandglass, either bulb A or B lies on top of the other and becomes the upper bulb. The other bulb becomes the lower bulb.

The sand drops from the upper bulb to the lower bulb at a rate of 1 gram per second. When the upper bulb no longer contains any sand, nothing happens.

Initially at time 0, bulb A is the upper bulb and contains a grams of sand; bulb B contains X-a grams of sand (for a total of X grams).

We will turn over the sandglass at time r_1,r_2,..,r_K. Assume that this is an instantaneous action and takes no time. Here, time t refer to the time t seconds after time 0.

You are given Q queries. Each query is in the form of (t_i,a_i). For each query, assume that a=a_i and find the amount of sand that would be contained in bulb A at time t_i.

Constraints

* 1≤X≤10^9
* 1≤K≤10^5
* 1≤r_1<r_2< .. <r_K≤10^9
* 1≤Q≤10^5
* 0≤t_1<t_2< .. <t_Q≤10^9
* 0≤a_i≤X (1≤i≤Q)
* All input values are integers.

Input

The input is given from Standard Input in the following format:


X
K
r_1 r_2 .. r_K
Q
t_1 a_1
t_2 a_2
:
t_Q a_Q


Output

For each query, print the answer in its own line.

Examples

Input

180
3
60 120 180
3
30 90
61 1
180 180


Output

60
1
120


Input

100
1
100000
4
0 100
90 100
100 100
101 100


Output

100
10
0
0


Input

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10


Output

19
52
91
10
58
42
100
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> P;
const long long LINF=1e18;
template<typename T>
void chmin(T &a,T b){
  if(a>b)a=b;
}
template<typename T>
void chmax(T &a,T b){
  if(a<b)a=b;
}

signed main(){
  int x,k;cin>>x>>k;
  vector<int> r(k); for(int &p:r)cin>>p;
  r.push_back(LINF);
  int q;cin>>q;
  vector<P> qu(q); for(P &p:qu){int t,a;cin>>t>>a;p=P(t,a);}
  int now=0,de=0,fl=1,low=0,high=x;
  int A=x;
  for(P _:qu){
    int t=_.first,a=_.second;
    while(now<=k&&r[now]<=t){
      de+=fl*(r[now]-(now>0?r[now-1]:0));
      chmin(high,x+de);
      chmax(low,de);
      A-=fl*(r[now]-(now>0?r[now-1]:0));chmax(A,0LL);chmin(A,x);
      fl*=-1;now++;
    }
    int ans,w=-de-fl*(t-(now?r[now-1]:0));
    if(a<low)ans=low+w;
    else if(a>high)ans=high+w;
    else ans=a+w;
    if(high<low)ans=A-fl*(t-(now?r[now-1]:0));
    chmax(ans,0LL);
    chmin(ans,x);
    cout<<ans<<endl;
  }
}
