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
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxM = 100000;
const int MaxN = 100000;

int x, m, n;
int rev[MaxM + 1];

int l, r, c;

inline void update(int delta)
{
	c += delta;

	if (l > x - c)
		l = r = x - c;
	else
		r = min(x - c, r);

	if (r < -c)
		l = r = -c;
	else
		l = max(l, -c);
}

int main()
{
	cin >> x >> m, r = x;
	for (int i = 1; i <= m; ++i)
		rev[i] = getint();

	cin >> n;

	int d = 0, k = 1, coef = -1;
	for (int i = 1; i <= n; ++i)
	{
		int t = getint(), a = getint();

		while (k <= m && rev[k] <= t)
		{
			update(coef * (rev[k] - d));
			d = rev[k++];
			coef = -coef;
		}
		update(coef * (t - d)), d = t;

		if (a <= l)
			printf("%d\n", l + c);
		else if (a <= r)
			printf("%d\n", a + c);
		else
			printf("%d\n", r + c);
	}

	return 0;
}