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;
typedef long long ll;
const int N = 3e5+1;
const int inf = 0x7ffffff;

int X,k,Q,ans[N];
struct nd{int t,a,id;}q[N];
bool cmp(nd a,nd b){return a.t<b.t;}
int up(int &x){if(x<0)x=0;if(x>X)x=X;}
int main()
{
	scanf("%d%d",&X,&k);
	for(int i=1;i<=k;++i)scanf("%d",&q[i].t);
	scanf("%d",&Q);
	for(int i=1;i<=Q;++i)
	scanf("%d%d",&q[i+k].t,&q[i+k].a),q[i+k].id=i;
	sort(q+1,q+Q+k+1,cmp);
	int a=0,b=X,l=0,r=X,tag=-1,pre=0;
	for(int i=1;i<=Q+k;++i)
	{
		if(q[i].id)
		{
			int ret=0,res=q[i].t-pre;
			if(q[i].a<=l)ret=a;
			else if(q[i].a>=r)ret=b;
			else ret=a-l+q[i].a;
			ret+=res*tag;
			up(ret);
			ans[q[i].id]=ret;
		}
		else
		{
			
			int v=q[i].t-pre;
			if(tag==-1)
			{
				if(v>a)
				{
					b-=v;l+=v-a;
					up(b);up(l);
					if(r<l)r=l;
					a=0;
				}
				else	a-=v,b-=v;
			}
			else
			{
				if(v+b>X)
				{
					a+=v;r-=b+v-X;
					up(a);up(r);
					if(r<l)l=r;
					b=X;
				}
				else	a+=v,b+=v;
			}
			tag*=-1;
			pre=q[i].t;
		}
	}
	for(int i=1;i<=Q;++i)printf("%d\n",ans[i]);
}