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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int x = sc.nextInt();
		int k = sc.nextInt();
		int[] r = new int[k];
		for(int i=0; i<k; i++){
			r[i] = sc.nextInt();
		}
			
		int q = sc.nextInt();
		int[] t = new int[q];
		int[] a = new int[q];
		for(int i=0; i<q; i++){
			t[i] = sc.nextInt();
			a[i] = sc.nextInt();
		}
		
		int currentA;
		boolean AisTop;
		int sub1, sub2;
		for(int i=0; i<q; i++){
			currentA = a[i];
			AisTop = true;
			for(int j=0; j<k; j++){
				if(j==0){
					sub1 = r[j];
					sub2 = t[i];
				}else{
					sub1 = r[j] - r[j-1];
					sub2 = t[i] - r[j-1];
				}
				if(AisTop){
					if(r[j]<t[i]){
						if(currentA<sub1){
							currentA=0;
						}else{
							currentA-=sub1;
						}
					}else{
						if(currentA<sub2){
							currentA=0;
							break;
						}else{
							currentA-=sub2;
							break;
						}
					}
				}else{
					if(r[j]<t[i]){
						if(x-currentA<sub1){
							currentA=x;
						}else{
							currentA+=sub1;
						}
					}else{
						if(x-currentA<sub2){
							currentA=x;
							break;
						}else{
							currentA+=sub2;
							break;
						}
					}
				}
				if(AisTop){
					AisTop = false;
				}else{
					AisTop = true;
				}
			}
			if(r[k-1]<t[i]){
				if(AisTop){
					if(currentA<t[i]-r[k-1]){
						currentA=0;
					}else{
						currentA-=t[i]-r[k-1];
					}
				}else{
					if(x-currentA<t[i]-r[k-1]){
						currentA=x;
					}else{
						currentA+=t[i]-r[k-1];
					}
				}
			}
			System.out.println(currentA);
		}
	}
}