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)

//#pragma warning disable

using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;

public class Test
{
    public static void Main()
    {
        var X = long.Parse(Console.ReadLine());
        var K = long.Parse(Console.ReadLine());
        var r = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();
        var Q = long.Parse(Console.ReadLine());

        // キャッシュ構築、時刻が１つズレるので注意
        var aRedMax = new long[K + 1];  // 赤文字
        var aRedMin = new long[K + 1];
        var aBlueMax = new long[K + 1];  // 青文字
        var aBlueMin = new long[K + 1];

        aRedMax[0] = X;
        aRedMin[0] = 0;
        aBlueMax[0] = X;
        aBlueMin[0] = 0;

        for (int i = 0; i < K; i++)
        {
            var timeElapsed = r[i] - (i == 0 ? 0 : r[i - 1]);

            aBlueMax[i + 1] = aBlueMax[i];
            aBlueMin[i + 1] = aBlueMin[i];

            if (i % 2 == 0)
            {
                // Aに入っている砂は減る
                aRedMax[i + 1] = aRedMax[i] - timeElapsed;
                aRedMin[i + 1] = aRedMin[i] - timeElapsed;

                if (aRedMin[i + 1] < 0)
                {
                    aBlueMin[i + 1] += (-aRedMin[i + 1]);
                    aRedMin[i + 1] = 0;
                }
            }
            else
            {
                // Aに入っている砂は増える
                aRedMax[i + 1] = aRedMax[i] + timeElapsed;
                aRedMin[i + 1] = aRedMin[i] + timeElapsed;

                if (aRedMax[i + 1] > X)
                {
                    aBlueMax[i + 1] -= (aRedMax[i + 1] - X);
                    aRedMax[i + 1] = X;
                }
            }
        }

        for (int i = 0; i < K + 1; i++)
        {
            //Console.WriteLine(aRedMax[i] + "," + aRedMin[i] + "," + aBlueMax[i] + "," + aBlueMin[i]);
        }
        
        for (int i = 0; i < Q; i++)
        {
            var line2 = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();
            var t = line2[0];
            var a = line2[1];

            var index = Array.BinarySearch<long>(r, t);

            if (index < 0) index = (~index);
            else index = index + 1;

            // aRedMax[index] をキャッシュとして用いる

            long timeElapsed = t - (index == 0 ? 0 : r[index - 1]);

            //Console.WriteLine(timeElapsed);
            //Console.WriteLine(aRedMax[index] + "," + aRedMin[index] + "," + aBlueMax[index] + "," + aBlueMin[index]);

            long result = -1;

            if (a < aBlueMin[index])
            {
                result = aRedMin[index];
            }
            else if (a > aBlueMax[index])
            {
                result = aRedMax[index];
            }
            else
            {
                result = a - aBlueMin[index] + aRedMin[index];
            }

            if (index % 2 == 0)
            {
                // Aの砂は減る
                result -= timeElapsed;
            }
            else
            {
                // Aの砂は増える
                result += timeElapsed;
            }

            if (result < 0) result = 0;
            if (result > X) result = X;

            Console.WriteLine(result);

        }




    }
}