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;
typedef long long ll;
int main(){
  ll x,k;cin>>x>>k;
  vector<ll> r(k+1);r[0]=0;for(ll i=1;i<=k;i++)cin>>r[i];
  ll d[k],u[k],da[k],ua[k];
  d[0]=0;
  da[0]=0;
  u[0]=x;
  ua[0]=x;
  for(ll i=1;i<=k;i++){
    ll t=r[i]-r[i-1];
    //cout<<t<<endl;
    if(i%2==0){
      if(da[i-1]+t>=x){
        d[i]=x;
        da[i]=x;
        u[i]=x;
        ua[i]=x;
      }
      else{
        da[i]=da[i-1]+t;
        d[i]=d[i-1];
        ua[i]=ua[i-1]+t;
        u[i]=u[i-1];
        if(ua[i]>=x){
          u[i]=u[i-1]-(ua[i]-x);
          ua[i]=x;
        }
      }
    }
    else{
      if(ua[i-1]-t<=0){
        d[i]=0;
        da[i]=0;
        u[i]=0;
        ua[i]=0;
      }
      else{
        da[i]=da[i-1]-t;
        d[i]=d[i-1];
        ua[i]=ua[i-1]-t;
        u[i]=u[i-1];
        if(da[i]<=0){
          d[i]=d[i]-da[i];
          da[i]=0;
        }
      }
    }
  }
  for(ll i=0;i<=k;i++){
    //cout<<d[i]<<" "<<da[i]<<" "<<u[i]<<" "<<ua[i]<<endl;
  }//return 0;
  ll q;cin>>q;
  for(ll i=0;i<q;i++){
    ll t,a;cin>>t>>a;
    ll e=upper_bound(r.begin(),r.end(),t)-r.begin();
    ll now;
    if(a<=d[e-1])now=da[e-1];
    else if(a>=u[e-1])now=ua[e-1];
    else now=a-d[e-1]+da[e-1];
    if(e%2==1){
      now-=t-r[e-1];
      now=max((ll)0,now);
    }
    else{
      now+=t-r[e-1];
      now=min(x,now);
    }
    cout<<now<<endl;
  }
}