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 100005
using namespace std;
int q,x,k;
int r[N];
struct node
{
	int a,t,op,y;
	friend bool operator < (const node &aa,const node &bb)
	{
		return aa.t<bb.t;
	}
}p[N*2];
int ans[N];
int main()
{
	scanf("%d%d",&x,&k);
	for(int i=1;i<=k;i++)scanf("%d",&r[i]),p[i].t=r[i],p[i].op=1;
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
	{
		p[i+k].op=2;p[i+k].y=i;
		scanf("%d%d",&p[i+k].t,&p[i+k].a);
	}
	sort(p+1,p+q+k+1);
	int t1=x,t2=0,ts=0;
	int now=1;
	for(int i=1;i<=k+q;i++)
	{
		if(now==1)
		{
			t1-=p[i].t-p[i-1].t;
			t2-=p[i].t-p[i-1].t;
			if(t1<0)t1=0;
			if(t2<0)t2=0;
		}
		else 
		{
			t1+=p[i].t-p[i-1].t;
			t2+=p[i].t-p[i-1].t;
			if(t1>x)ts+=t1-x,t1=x;
			if(t2>x)t2=x;
			if(ts>x)ts=x;
		}
		if(p[i].op==1)now^=1;
		else 
		{
			int tmp=x-p[i].a;
			int tt=max(0,min(t1,t1-tmp+ts));
			if(tt<t2)tt=t2;
			ans[p[i].y]=tt;
		}
	}
	for(int i=1;i<=q;i++)
	{
		printf("%d\n",ans[i]);
	}
	return 0;
}
