Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

180
3
60 120 180
3
30 90
61 1
180 180

Sample Output 1

60
1
120

In the first query, 30 out of the initial 90 grams of sand will drop from bulb A, resulting in 60 grams.
In the second query, the initial 1 gram of sand will drop from bulb A, and nothing will happen for the next 59 seconds. Then, we will turn over the sandglass, and 1 second after this, bulb A contains 1 gram of sand at the time in question.

Sample Input 2

100
1
100000
4
0 100
90 100
100 100
101 100

Sample Output 2

100
10
0
0

In every query, the upper bulb initially contains 100 grams, and the question in time comes before we turn over the sandglass.

Sample Input 3

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10

Sample Output 3

19
52
91
10
58
42
100

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int action(int x, int k, int* r, int t, int a) {
  int i;
  int highest = k;
  int lowest = 0;
  if (r[k] < t) {
    lowest = k;
  } else {
    while (highest > lowest+1) {
      int mid = (highest+lowest)/2;
      if (r[mid] > t) {
	highest = mid;
      } else {
	lowest = mid;
      }
    }
  }
  int r_last = lowest;
  /*printf("%d\n", r_last);*/
  int up = a;
  int down = x-a;
  for (i = 1; i <= r_last; i++) {
    int time = r[i] - r[i-1];
    down = fmax(up - time, 0);
    up = x-down;
    /*printf("%d %d\n", up, down);*/
  }
  up = fmax(up-(t-r[r_last]), 0);
  down = x-up;
  if (r_last % 2 == 0) {
    return up;
  } else {
    return down;
  }
}

int main() {
  int x, k, q, j;
  scanf("%d", &x);
  scanf("%d", &k);
  int r[100010] = {};
  for (j = 1; j < k+1; j++) {
    scanf("%d", &r[j]);
  }
  scanf("%d", &q);
  int t[100010], a[100010], ans[100010];
  for (j = 1; j < q+1; j++) {
    scanf("%d %d", &t[j], &a[j]);
    ans[j] = action(x, k, r, t[j], a[j]);
  }
  for (j = 1; j < q+1; j++) {
    printf("%d\n", &ans[j]);
  }
  return 0;
}