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 <cstdio>
#include <algorithm>
const int maxn = 1e5 + 5;
using namespace std;
struct QNode{
	int t, a, res, id;
	bool operator< (QNode x)const{
		return t < x.t || (t == x.t && a < x.a);
	}
}q[maxn]; int X, n, Q, r[maxn], opt = -1;
bool cmpid(QNode x, QNode y){
	return x.id < y.id;
}
int main(){
	scanf("%d%d", &X, &n); for (int i = 1; i <= n; i++) scanf("%d", r + i);
	scanf("%d", &Q); for (int i = 1; i <= Q; i++) scanf("%d%d", &q[i].t, &q[i].a), q[i].id = i;
	sort(q + 1, q + Q + 1);
	int lr = 0, lq = 1, u = X, d = 0, S = 0;
	while (lq <= Q){
		if (r[lr + 1] < q[lq].t && lr < n){
			++lr; int t = opt * (r[lr] - r[lr - 1]);
			d = max(0, min(X, d + t));
			u = max(0, min(X, u + t));
			S += t; opt *= -1;
		}
		else{
			q[lq].res = max(d, min(u, S + q[lq].a));
			q[lq].res = min(X, max(0, q[lq].res + opt * (q[lq].t - r[lr])));
			++lq;
		}
	} sort(q + 1, q + Q + 1, cmpid);
	for (int i = 1; i <= Q; i++) printf("%d\n", q[i].res);
	return 0;
}