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;
const int N=2e5+10;
struct opt{int t,id;}Q[N];
bool cmp(opt a,opt b){return a.t<b.t;}
int X,n,q,a[N],ans[N],pre,suf,pre_V,suf_V;
int last,sig;
int main()
{
	scanf("%d%d",&X,&n);
	for (int i=1;i<=n;i++) scanf("%d",&Q[i].t);
	scanf("%d",&q);
	for (int i=1;i<=q;i++) scanf("%d%d",&Q[n+i].t,&a[i]),Q[n+i].id=i;
	sort(Q+1,Q+n+q+1,cmp);
	pre=pre_V=0;suf=suf_V=X;sig=-1;
	for (int i=1;i<=n+q;i++){
		int d=sig*(Q[i].t-last);last=Q[i].t;
		pre_V+=d;suf_V+=d;
		if (pre_V<0) pre=min(pre-pre_V,X),pre_V=0;
		pre_V=min(pre_V,X);
		if (suf_V>X) suf=max(suf+X-suf_V,0),suf_V=X;
		suf_V=max(suf_V,0);
		if (pre>=suf) pre=X,suf=0;
		if (Q[i].id){
			int id=Q[i].id;
			if (a[id]<=pre) ans[id]=pre_V;else
			if (a[id]>=suf) ans[id]=suf_V;else
			ans[id]=pre_V+a[id]-pre;
		}
		else sig=-sig;
	}
	for (int i=1;i<=q;i++) printf("%d\n",ans[i]);
	return 0;
}