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
signed main(){
  int x,k;
  cin>>x>>k;
  int r[k];
  for(int i=0;i<k;i++) cin>>r[i];
  int q;
  cin>>q;
  int t[q],a[q];
  for(int i=0;i<q;i++) cin>>t[i]>>a[i];

  int L=0,R=x;//[L,R]
  int lb[k],ub[k];
  fill_n(lb,k,0);
  fill_n(ub,k,x);
  
  int j=0,p=0,d=0;
  int lv=0,uv=x;
  for(int i=0;i<q;i++){
    while(j<k&&r[j]<=t[i]){
      if(j%2==0) p+=r[j]-d;
      else p-=r[j]-d;
      if(j%2==0){
	lv=max(lv-(r[j]-d),0LL);
	uv=max(uv-(r[j]-d),0LL);
      }else{
	lv=min(lv+(r[j]-d),x);
	uv=min(uv+(r[j]-d),x);
      }
      
      L=max(L,p);
      R=min(R,x+p);
      lb[j]=L;
      ub[j]=R;
      d=r[j++];
      
      
    }
    //cout<<i<<" "<<j<<":"<<p<<endl;
    //cout<<L<<" "<<R<<endl;
    if(L<=a[i]&&a[i]<=R){
      int c=a[i]-p;
      if(j%2==0) c=max(c-(t[i]-d),0LL);
      else c=min(c+(t[i]-d),x);
      cout<<c<<endl;
    }else if(a[i]<L){
      int c=lv;
      if(j%2==0) c=max(c-(t[i]-d),0LL);
      else c=min(c+(t[i]-d),x);
      cout<<c<<endl;
    }else if(R<a[i]){
      int c=uv;
      if(j%2==0) c=max(c-(t[i]-d),0LL);
      else c=min(c+(t[i]-d),x);
      cout<<c<<endl;
    }else{
      cout<<"?"<<endl;
    }
  }
  return 0;
}
