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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

int main(){
  ll X; int K; cin>>X>>K;
  set<pair<ll,ll> > st;
  rep(i,K){
    ll t; cin>>t;
    st.insert(make_pair(t,-1));
  }
  int Q; cin>>Q;
  rep(i,Q){
    ll t,a; cin>>t>>a;
    st.insert(make_pair(t,a));
  }
  bool inc=0;
  ll pre=0, a=0,b=X,c=0;
  for(auto s:st){
    ll t=s.first, x=s.second;
    ll dt=t-pre;
    if(inc){
      c+=dt;
      if(b+c>X) b=X-c;
      if(a>b) a=b;
    }else{
      c-=dt;
      if(a+c<0) a=-c;
      if(a>b) b=a;
    }
    if(x==-1) inc=!inc;
    else cout<<max(a,min(b,x))+c<<endl;
    pre=t;
  }
  return 0;
}
