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<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
//#include<iostream>
using namespace std;

#define LL long long
int n,m;
#define maxn 200011
struct event
{
	int t,a,id;
	bool operator < (const event &b) const {return t<b.t;}
}e[maxn];int le=0;
LL X,ans[maxn];
int main()
{
	scanf("%lld%d",&X,&n);
	for (int i=1;i<=n;i++) le++,scanf("%d",&e[le].t),e[le].id=e[le].a=0;
	scanf("%d",&m);
	for (int i=1;i<=m;i++) le++,scanf("%d%d",&e[le].t,&e[le].a),e[le].id=i;
	sort(e+1,e+1+le);
//	for (int i=1;i<=le;i++) cout<<e[i].t<<' '<<e[i].a<<' '<<e[i].id<<endl;
	LL Min=0,Max=X,pre=0;
	for (int i=1,sig=-1,last=0;i<=le;i++)
	{
		LL now=(e[i].t-last)*sig;
		if (e[i].id)
		{
			LL v=e[i].a+pre;
			if (v<Min) v=Min;
			if (v>Max) v=Max;
			v+=now;
			if (v<0) v=0;
			if (v>X) v=X;
			ans[e[i].id]=v;
		}
		else
		{
			Min+=now;Max+=now;
			if (Min<0) Min=0;if (Max<0) Max=0;
			if (Min>X) Min=X;if (Max>X) Max=X;
			pre+=now;sig*=-1;last=e[i].t;
		}
	}
	for (int i=1;i<=m;i++) printf("%lld\n",ans[i]);
	return 0;
}