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 <iostream>
#include <vector>
#include <queue>

using namespace std;

int main(){
  int X, K, Q;
  int r[100010];
  cin >> X;
  cin >> K;
  for(int i = 0; i < K; i++)  cin >> r[i];
  cin >> Q;
  r[K] = 1e9 + 1;

  int x, y, z;
  int t, a;
  int tim = 0, dir = 1, rin = 0;
  x = 0; y = X; z = 0;
  for(int i = 0; i < Q; i++){
    cin >> t >> a;
    while(1){
      int tt = min(r[rin], t);
      if(dir == 1){
        z = max(z-(tt - tim), -X);
        x = max(x, -z);
        y = max(x, y);
      }
      else{
        z = min(z+(tt - tim), X);
        y = min(y, X-z);
        x = min(x, y);
      }

      if(tt == r[rin]){
        tim = r[rin++];
        dir = 1 - dir;
      }
      if(tt == t){
        tim = t;
        break;
      }
    }
    if(a <= x) cout << max(0, min(x + z, X)) << endl;
    else if(a >= y) cout << max(0, min(y + z, X)) << endl;
    else  cout << max(0, min(a + z, X)) << endl;

    //cout << x << " " << y << " " << z << endl;
    //cout << dir << endl;

  }

  return 0;
}
