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 ll long long
using namespace std;
int x,k,r[100005],t,num;
inline int read()
{int x=0;
char c=getchar();
while (c<'0'||c>'9') c=getchar();
while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
return x;
}
int main (){
	int i,j,q;
	x=read();k=read();
	for (i=1;i<=k;i++)
	{r[i]=read();}
	int a,minn=0,maxn=x,pos=1,ansmin=0,ansmax=x,num=0;
	q=read();
	while (q--)
	{t=read();a=read();
	while (pos<=k&&r[pos]<t)
	{ansmin-=(r[pos]-r[pos-1]);
	if (ansmin<0) ansmin=0;
	ansmin=x-ansmin;
	ansmax-=(r[pos]-r[pos-1]);
	if (ansmax<0) ansmax=0;
	ansmax=x-ansmax;
	if (pos&1) {num-=(r[pos]-r[pos-1]);}
	else {num+=(r[pos]-r[pos-1]);}
	if (num<0) {minn=max(minn,-num);}
	if (num>0) {maxn=min(maxn,x-num);}
	pos++;
	}
	//printf ("%d %d\n",ansmin,ansmax);
	int tpans;
	if (a<=minn) {tpans=ansmin;}
	if (a>=maxn) {tpans=ansmax;}
	if (a>minn&&a<maxn) {tpans=ansmin+(a-minn)*((ansmin>ansmax)?-1:1);}
	tpans-=(t-r[pos-1]);
	if (tpans<0) tpans=0;
	if ((pos-1)&1) {tpans=x-tpans;}
	printf ("%d\n",tpans);
	}
	return 0;
}
	
	