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 Add(x,y) x=mymax(0,mymin(X,x+y))
using namespace std;

namespace IO
{
	const int S=(1<<20)+5;
	//Input Correlation
	char buf[S],*H,*T;
	inline char Get()
	{
		if(H==T) T=(H=buf)+fread(buf,1,S,stdin);
		if(H==T) return -1;return *H++;
	}
	inline int read()
	{
		int x=0;char c=Get();
		while(!isdigit(c)) c=Get();
		while(isdigit(c)) x=x*10+c-'0',c=Get();
		return x;
	}
	//Output Correlation
	char obuf[S],*oS=obuf,*oT=oS+S-1,c,qu[55];int qr;
	inline void flush(){fwrite(obuf,1,oS-obuf,stdout);oS=obuf;}
	inline void putc(char x){*oS++ =x;if(oS==oT) flush();}
	template <class I>inline void print(I x)
	{
		if(!x) putc('0');
		if(x<0) putc('-'),x=-x;
		while(x) qu[++qr]=x%10+'0',x/=10;
		while(qr) putc(qu[qr--]);
	}
}

inline int mymax(const int &x,const int &y){return x>y?x:y;}
inline int mymin(const int &x,const int &y){return x<y?x:y;}

using namespace IO;
const int N=100010;
int t[N],a[N],r[N];
int X,n,Q;

int main()
{
	X=read();n=read();
	for(int i=1;i<=n;i++) r[i]=read();
	Q=read();
	for(int i=1;i<=Q;i++) t[i]=read(),a[i]=read();
	int down=0,c=0,up=X,fg=-1,p1=0,p2=1,ad=0;
	while(p2<=Q)
	{
		if(t[p2]>r[p1+1]&&p1<n)
		{
			p1++;
			ad=fg*(r[p1]-r[p1-1]);
			Add(down,ad);Add(up,ad);
			c+=ad;fg=-fg;
		}
		else
		{
			int len=t[p2]-r[p1];
			int ans=mymax(down,mymin(up,a[p2]+c));
			Add(ans,fg*len);
			print(ans);putc('\n');
			p2++;
		}
	}
	flush();
	return 0;
}