#include <algorithm>
#include <cstdio>
#define calc(l, r, t) std::min(l, std::max(r, t))
int seq[100005]; 
int main()
{
	// freopen("ARC082-F.in", "r", stdin);
	int x, n; 
	scanf("%d%d", &x, &n); 
	for (int i = 0; i < n; i++)
		scanf("%d", seq + i); 
	int q, idx = 0, coef = -1, l = x, r = 0, c = 0; 
	scanf("%d", &q); 
	while (q--)
	{
		int t, a; 
		scanf("%d%d", &t, &a); 
		while (idx < n && t >= seq[idx])
		{
			int dt = coef * (seq[idx] - (idx ? seq[idx - 1] : 0)); 
			l = calc(x, 0, l + dt); 
			r = calc(x, 0, r + dt); 
			c += dt; 
			idx++; 
			coef *= -1; 
		}
		int dt = coef * (t - (idx ? seq[idx - 1] : 0)); 
		printf("%d\n", calc(x, 0, calc(l, r, a + c) + dt));
	}
	return 0; 
}

### Prompt

Generate a Cpp solution to the following problem:

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.

Examples

Input

180
3
60 120 180
3
30 90
61 1
180 180


Output

60
1
120


Input

100
1
100000
4
0 100
90 100
100 100
101 100


Output

100
10
0
0


Input

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10


Output

19
52
91
10
58
42
100

### Response

```cpp
#include <algorithm>
#include <cstdio>
#define calc(l, r, t) std::min(l, std::max(r, t))
int seq[100005]; 
int main()
{
	// freopen("ARC082-F.in", "r", stdin);
	int x, n; 
	scanf("%d%d", &x, &n); 
	for (int i = 0; i < n; i++)
		scanf("%d", seq + i); 
	int q, idx = 0, coef = -1, l = x, r = 0, c = 0; 
	scanf("%d", &q); 
	while (q--)
	{
		int t, a; 
		scanf("%d%d", &t, &a); 
		while (idx < n && t >= seq[idx])
		{
			int dt = coef * (seq[idx] - (idx ? seq[idx - 1] : 0)); 
			l = calc(x, 0, l + dt); 
			r = calc(x, 0, r + dt); 
			c += dt; 
			idx++; 
			coef *= -1; 
		}
		int dt = coef * (t - (idx ? seq[idx - 1] : 0)); 
		printf("%d\n", calc(x, 0, calc(l, r, a + c) + dt));
	}
	return 0; 
}

```