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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define pp pair<int,int>
#define Pi acos(-1)
#define pq priority_queue
#define mz 1000000007
 
using namespace std;
 
int x,k,q,t;
int r[100005];
int d[100005];
int lu[100005],ld[100005],ru[100005],rd[100005];
 
int main()
{
    int a,ans,nowlu,nowld,nowru,nowrd,temp=0;
    scanf("%d",&x);
    scanf("%d",&k);
    for(int i=1;i<=k;i++)
        scanf("%d",&r[i]);
    for(int i=1;i<=k;i++)
        d[i]=min(r[i]-r[i-1],x);
    lu[0]=ru[0]=x;
    ld[0]=rd[0]=0;
    r[k+1]=mz;
    for(int i=1;i<=k;i++)
    {
        if(i%2)
        {
            nowld=d[i];
            nowlu=x;
            nowrd=0;
            nowru=x-d[i];
        }
        else
        {
            nowld=0;
            nowlu=x-d[i];
            nowrd=d[i];
            nowru=x;
        }
        if(ru[i-1]<=nowld || rd[i-1]>=nowlu)
        {
            if(i%2)
            {
                ld[i]=x;
                lu[i]=x;
                rd[i]=0;
                ru[i]=0;
            }
            else
            {
                ld[i]=0;
                lu[i]=0;
                rd[i]=x;
                ru[i]=x;
            }
        }
        else
        {
            if(rd[i-1]<=nowld)
            {
                ld[i]=ld[i-1]+nowld-rd[i-1];
                rd[i]=nowrd;
            }
            else
            {
                ld[i]=ld[i-1];
                rd[i]=nowrd-nowld+rd[i-1];
            }
            if(ru[i-1]>=nowlu)
            {
                lu[i]=lu[i-1]+nowlu-ru[i-1];
                ru[i]=nowru;
            }
            else
            {
                lu[i]=lu[i-1];
                ru[i]=nowru-nowlu+ru[i-1];
            }
        }
    }
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d%d",&t,&a);
        while(r[temp+1]<=t)
            temp++;
        if(a>=ld[temp] && a<=lu[temp])
            ans=a-ld[temp]+rd[temp];
        else
        {
            if(a<=ld[temp])
                ans=rd[temp];
            else
                ans=ru[temp];
        }
        if(temp%2)
            ans=min(x,ans+(t-r[temp]));
        else
            ans=max(0,ans-(t-r[temp]));
        printf("%d\n",ans);
    }
    return 0;
}