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 lli;
typedef double lld;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<double> vdl;
typedef vector<vector<lli>> mat;
typedef vector<vdl> mad;
typedef unordered_map<lli,unordered_map<lli,lli>> graph;
typedef complex<double> cmp;
typedef vector<cmp> vcl;

lli x,k;
vll r;
lli q;
vll t,a;

int main(){
  cin >> x >> k;
  r = vll(k+1);
  for(lli i = 1;i <= k;i++) cin >> r[i];
  cin >> q;
  t = a = vll(q);
  for(lli i = 0;i < q;i++) cin >> t[i] >> a[i];
  lli l,u,s;
  l = 0;u = x;s = 0;
  for(lli i = 0,j = 0;i < q;i++){
    while(j+1 <= k && r[j+1] < t[i]){
      if(j%2){
        lli d = min(r[j+1]-r[j],x-l-s);
        u = min(u,x-s-d);
        s += d;
      }else{
        lli d = max(r[j]-r[j+1],-u-s);
        l = max(l,-s-d);
        s += d;
      }
      j++;
    }
    if(j%2){
      lli d = min(t[i]-r[j],x-l-s);
      u = min(u,x-s-d);
      s += d;
    }else{
      lli d = max(r[j]-t[i],-u-s);
      l = max(l,-s-d);
      s += d;
    }
    r[j] = t[i];
    if(a[i] <= l) cout << l+s << endl;
    else if(a[i] >= u) cout << u+s << endl;
    else cout << a[i]+s << endl;
  }


  return 0;

}
