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>
#define N 300005
using namespace std;
int X,n,m,q,op,ans[N];
struct node{
	int t,a,num;
	node(int _t=0,int _a=0,int _num=0){ t=_t,a=_a,num=_num;}
}a[N];
bool cmp(const node &p,const node &q){ return p.t<q.t||(p.t==q.t&&p.num<q.num);}
int main()
{
	int i,x,y,p,l,r,v;
	scanf("%d",&X);
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		scanf("%d",&x),a[++m]=node(x,0,0);
	scanf("%d",&q);
	for(i=1;i<=q;i++)
		scanf("%d %d",&x,&y),a[++m]=node(x,y,i);
	sort(a+1,a+m+1,cmp);
	op=-1,p=l=v=0,r=X;
	for(i=1;i<=m;i++){
		y=(a[i].t-p)*op;
		if(a[i].num!=0){
			x=a[i].a+v;
			x=min(x,r);
			x=max(x,l);
			x+=y;
			x=min(x,X);
			x=max(x,0);
			ans[a[i].num]=x;
		  }
		else{
			r+=y,l+=y,v+=y;
			r=min(r,X),r=max(r,0);
			l=min(l,X),l=max(l,0);
			p=a[i].t,op=-op;
		  }
	  }
	for(i=1;i<=q;i++)
		printf("%d\n",ans[i]);
	return 0;
}
