#include<iostream>
#define m(a) min(X,max(0,a))
using namespace std;

int r[200000];
int X,K,Q,T,A,s,l,k,t,x,f=-1;

int main()
{
  cin>>X>>K;
  for(;l<K;)cin>>r[l++];

  l = X;
  
  for(cin >> Q; Q--;) {
    cin >> T >> A;
    for(; k<K&T>=r[k]; ) {
      int d = f * (r[k] - t);
      x += d;
      s=m(s+d);
      l=m(l+d);
      t = r[k];
      k++;
      f *= -1;
    }
    cout << m(min(l, max(s, A + x)) + f * (T-t)) << endl;
  }

  return 0;
}
### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include<iostream>
#define m(a) min(X,max(0,a))
using namespace std;

int r[200000];
int X,K,Q,T,A,s,l,k,t,x,f=-1;

int main()
{
  cin>>X>>K;
  for(;l<K;)cin>>r[l++];

  l = X;
  
  for(cin >> Q; Q--;) {
    cin >> T >> A;
    for(; k<K&T>=r[k]; ) {
      int d = f * (r[k] - t);
      x += d;
      s=m(s+d);
      l=m(l+d);
      t = r[k];
      k++;
      f *= -1;
    }
    cout << m(min(l, max(s, A + x)) + f * (T-t)) << endl;
  }

  return 0;
}
```