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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		long x = long.Parse(Console.ReadLine());
		int k = int.Parse(Console.ReadLine());
    long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
    long[] times = new long[k+1];
    for(int i = 0; i < k; i++) times[i+1] = nums[i];
    int reverseCount = 0;
    int q = int.Parse(Console.ReadLine());
    long sum = 0;//初期からのAの増減
    long Azero = 0;//初期0の時のAの変動
    long Bzero = x;//初期xの時
    long AzeroMax = 0;//Aが0になるAの最大値
    long BzeroMin = x;//AがxになるBの最小値

    for(int i = 0; i < q; i++)
    {
      string[] input = Console.ReadLine().Split(' ');
      long t = long.Parse(input[0]);
      long a = long.Parse(input[1]);

      if(reverseCount < k)
      {
        while(times[reverseCount+1] < t)
        {
          long tt = times[reverseCount+1];
          reverseCount++;
          if(reverseCount % 2 == 1)//Aが減る
          {
            sum -= tt;
            AzeroMax = Math.Max(AzeroMax, -sum);
          
            Azero -= tt;
            if(Azero < 0) Azero = 0;
            else if(Azero > x) Azero = x;
            Bzero -= tt;
            if(Bzero < 0) Bzero = 0;
            else if(Bzero > x) Bzero = x;
          }else
          {
            sum += tt;
            BzeroMin = Math.Min(BzeroMin, sum);
        
            Azero += tt;
            if(Azero < 0) Azero = 0;
            else if(Azero > x) Azero = x;
            Bzero += tt;
            if(Bzero < 0) Bzero = 0;
            else if(Bzero > x) Bzero = x;
          }
          if(reverseCount == k) break;
        }  
      }
      if(a <= AzeroMax) Console.WriteLine(Azero);
      else if(a >= BzeroMin) Console.WriteLine(Bzero);
      else
      {
        long answer = a + sum;
        if(answer < 0) answer = 0;
        else if(answer > x) answer = x;

        if(reverseCount % 2 == 1)
        {
          answer += t - times[reverseCount];
        }else
        {
          answer -= t - times[reverseCount];
        }

        if(answer < 0) answer = 0;
        else if(answer > x) answer = x;
        Console.WriteLine(answer);
      }
    }
	}
}