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 <cstdio>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>

using ll = long long int;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

ll v[100100];

int main(void){
	int n,q,i,j=1;
	ll ans,l,r,s=0ll,a,x,t,tmpl,tmpr;
	std::cin >> x; l = 0ll; r = x;
	std::cin >> n; ++n;
	for(i=1; i<n; ++i){
		std::cin >> v[i];
	}
	v[n] = INFLL;
	std::cin >> q;
	for(i=0; i<q; ++i){
		std::cin >> t >> a;
		if(!(t<v[j])){
			if(j%2){
				tmpl = (v[j]-v[j-1]); tmpr = x;
				tmpl -= s; tmpr -= s;
				l = std::max(tmpl,l); r = std::min(tmpr,r);
				s += v[j-1]-v[j];
				if(r<l){
					r = l;
				}
			}else{
				tmpl = 0ll; tmpr = x-(v[j]-v[j-1]);
				tmpl -= s; tmpr -= s;
				l = std::max(tmpl,l); r = std::min(tmpr,r);
				s += v[j]-v[j-1];
				if(l>r){
					l = r;
				}
			}
			// std::cout << l << " " << r << std::endl;
			++j;
		}
		a = std::min(std::max(a,l),r);
		if(j%2){
			ans = std::max(a+s-(t-v[j-1]),0ll);
		}else{
			ans = std::min(a+s+(t-v[j-1]),x);
		}
		std::cout << ans << std::endl;
	}
	return 0;
}