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 <bits/stdc++.h>

using namespace std;

#define ll long long
#define up(i,j,n)		for (int i = j; i <= n; i++)
#define down(i,j,n)	for (int i = j; i >= n; i--)
#define cmax(a,b)		a = max (a, b)
#define cmin(a,b)		a = min (a, b)

const int MAXN = 1e5 + 5;
const int oo = 0x3f3f3f3f;

int N, r[MAXN], X, Q;

int main(){
	scanf("%d%d", &X, &N);
	up (i, 1, N) scanf("%d", &r[i]);
	scanf("%d", &Q);
	int cur = 1, c = 0, lower = 0, upper = X, tt = -1;
	while (Q--) {
		int t, a;
		scanf("%d%d", &t, &a);
		while (r[cur] <= t && cur <= N) {
			int d = r[cur] - r[cur - 1];
			c += tt * d;
			if (lower + c < 0) {
				if (c >= -X) lower = -c;
				else {
					lower = X;
					c = -X;
				}
			}else if (lower + c > X) {
				if (c <= X) lower = X - c;
				else {
					lower = 0;
					c = X;
				}
			}
			if (upper + c < 0) {
				if (c >= -X) upper = -c;
				else {
					upper = X;
					c = -X;
				}
			}else if (upper + c > X) {
				if (c <= X) upper = X - c;
				else {
					upper = 0;
					c = X;
				}
			}
			tt *= -1;
			cur++;
		}
		int d = t - r[cur - 1];
		cmax (a, lower);
		cmin (a, upper);
		a += c + tt * d;
		if (a < 0) a = 0;
		if (a > X) a = X;
		printf("%d\n", a);
	}
	return 0;
}
