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 maxn=2e5+10;

int S,n,m,way=-1;
int La,Lp,Ra,Rp,Ans[maxn];

struct events{
	int t,a,id;
	events() {}
	events(int t,int a,int id):t(t),a(a),id(id) {}
	bool operator < (const events &e) const {return t<e.t;}
}e[maxn];

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i,t,a;
	scanf("%d",&S);
	scanf("%d",&n);
	for (i=1;i<=n;i++)
	{
		scanf("%d",&t);
		e[i]=events(t,-1,0);
	}
	scanf("%d",&m);
	for (i=1;i<=m;i++)
	{
		scanf("%d%d",&t,&a);
		e[n+i]=events(t,a,i);
	}
	sort(e+1,e+n+m+1);

	La=0; Lp=0; Ra=S; Rp=S;
	for (i=1;i<=n+m;i++)
	{
		Lp+=way*(e[i].t-e[i-1].t);
		Rp+=way*(e[i].t-e[i-1].t);
		if (Lp<0)
		{
			if (Rp<0) La=Ra=S,Lp=Rp=0;
			else La-=Lp,Lp=0;
		}
		if (Rp>S)
		{
			if (Lp>S) La=Ra=0,Lp=Rp=S;
			else Ra-=Rp-S,Rp=S;
		}
		if (e[i].a==-1) way=-way;
		else
		{
			int tmp=Lp+e[i].a-La;
			tmp=min(tmp,Rp);
			tmp=max(tmp,Lp);
			Ans[e[i].id]=tmp;
		}
	}
	for (i=1;i<=m;i++)
		printf("%d\n",Ans[i]);
}
