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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long

int max(int a, int b){
	if(a > b){
		return a;
	}
	else{
		return b;
	}
}

int min(int a, int b){
	if(a < b){
		return a;
	}
	else{
		return b;
	}
}

signed main(){
	int X, K, Q, i;
	scanf("%lld%lld", &X, &K);
	int *r = (int *)malloc(sizeof(int) * (K + 1));
	r[0] = 0;
	for(i = 1; i <= K; i++){
		scanf("%lld", &r[i]);
	}
	scanf("%lld", &Q);
	int *t = (int *)malloc(sizeof(int) * Q);
	int *a = (int *)malloc(sizeof(int) * Q);
	for(i = 0; i < Q; i++){
		scanf("%lld%lld", &t[i], &a[i]);
	}
	int *termsd = (int *)malloc(sizeof(int) * (K + 1));
	int *termsu = (int *)malloc(sizeof(int) * (K + 1));
	int *Ad = (int *)malloc(sizeof(int) * (K + 1));
	int *An = (int *)malloc(sizeof(int) * (K + 1));
	int *Au = (int *)malloc(sizeof(int) * (K + 1));
	termsd[0] = 0;
	termsu[0] = X;
	Ad[0] = 0;
	An[0] = 0;
	Au[0] = X;
	for(i = 1; i <= K; i++){
		if(i % 2 == 1){
			Ad[i] = min(X, max(0, Ad[i - 1] + r[i - 1] - r[i]));
			An[i] = An[i - 1] + r[i - 1] - r[i];
			Au[i] = min(X, max(0, Au[i - 1] + r[i - 1] - r[i]));
			termsd[i] = max(termsd[i - 1], -An[i]);
			termsu[i] = min(termsu[i - 1], X - An[i]);
		}
		else{
			Ad[i] = min(X, max(0, Ad[i - 1] - r[i - 1] + r[i]));
			An[i] = An[i - 1] - r[i - 1] + r[i];
			Au[i] = min(X, max(0, Au[i - 1] - r[i - 1] + r[i]));
			termsd[i] = max(termsd[i - 1], -An[i]);
			termsu[i] = min(termsu[i - 1], X - An[i]);
		}
	}
	int left, half, right;
	int ans1, ans2, j;
	for(i = 0; i < Q; i++){
		left = 0;
		right = K + 1;
		while(right - left > 1){
			half = (left + right) / 2;
			if(r[half] < t[i]){
				left = half;
			}
			else{
				right = half;
			}
		}
		if(left % 2 == 0){
			if(a[i] <= termsd[left]){
				ans1 = max(0, Ad[left] + r[left] - t[i]);
			}
			else if(a[i] < termsu[left]){
				ans1 = max(0, a[i] + An[left] + r[left] - t[i]);
			}
			else{
				ans1 = max(0, Au[left] + r[left] - t[i]);
			}
		}
		else{
			if(a[i] <= termsd[left]){
				ans1 = min(X, Ad[left] - r[left] + t[i]);
			}
			else if(a[i] < termsu[left]){
				ans1 = min(X, a[i] + An[left] - r[left] + t[i]);
			}
			else{
				ans1 = min(X, Au[left] - r[left] + t[i]);
			}
		}
		printf("%lld\n", ans1);
/*		ans2 = a[i];
		for(j = 1; r[j] <= t[i] && j <= K; j++){
			if(j % 2 == 1){
				ans2 = max(0, ans2 + r[j - 1] - r[j]);
			}
			else{
				ans2 = min(X, ans2 - r[j - 1] + r[j]);
			}
		}
		if(j % 2 == 1){
			ans2 = max(0, ans2 + r[j - 1] - t[i]);
		}
		else{
			ans2 = min(X, ans2 - r[j - 1] + t[i]);
		}
		printf("%lld\n", ans2);
*/	}
	return 0;
}