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: Accepted)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>

using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;

int main()
{
	ll x; int k;
	cin>>x>>k;
	ll r[100001], ra[100001];
	r[0]=0; ra[0]=0;
	for(int i=1; i<=k; i++){
		cin>>r[i];
		if(i%2) ra[i]=ra[i-1]+r[i]-r[i-1];
		else ra[i]=ra[i-1]-r[i]+r[i-1];
	}
	int q;
	cin>>q;
	ll a[100000];
	ll t[100000];
	for(int i=0; i<q; i++){
		cin>>t[i]>>a[i];
	}
	ll x1=0, x2=x;
	ll ans0=-1;
	int j=0;
	for(int i=0; i<=k; i++){
		while(j<q && (i==k || t[j]<r[i+1])){
			ll t1=t[j]-r[i], ans;
			if(ans0>=0){
				if(i%2) ans=min(x, ans0+t1);
				else ans=max(0ll, ans0-t1);
			}else if(x1>a[j]){
				if(i%2) ans=min(x, x1-ra[i]+t1);
				else ans=max(0ll, x1-ra[i]-t1);
			}else if(x2<a[j]){
				if(i%2) ans=min(x, x2-ra[i]+t1);
				else ans=max(0ll, x2-ra[i]-t1);
			}else{
				if(i%2) ans=min(x, a[j]-ra[i]+t1);
				else ans=max(0ll, a[j]-ra[i]-t1);
			}
          cout<<ans<<endl;
			j++;
		}
		if(i==k) break;
		if(ans0>=0){
			if(i%2) ans0=min(x, ans0+r[i+1]-r[i]);
			else ans0=max(0ll, ans0-r[i+1]+r[i]);
		}else if(i%2){
			if(x+ra[i+1]<x2){
				x2=x+ra[i+1];
			}
			if(x1>=x2){
				ans0=x;
			}
		}else{
			if(ra[i+1]>x1){
				x1=ra[i+1];
			}
			if(x1>=x2){
				ans0=0;
			}
		}
	}
    return 0;
}