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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline int read(){
	int n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int maxn=1e5+5;
int T,i,j,now,r[maxn],Q,t,x,mx,mi,c,k,ans,lami,lamx;
bool bz;
int main(){
	T=read(),k=read();
	fo(i,1,k) r[i]=read();
	Q=read();
	mi=0,mx=T,c=0;
	now=1,bz=1;
	for(;Q--;){
		t=read(),x=read();
		for(;now<=k&&t>r[now];now++,bz^=1){
			int len=r[now]-r[now-1];
			if (bz) lami=mi,mi=max(mi,min(mx,len-c)),c=max(0,lami+c-len)-mi;
			else lamx=mx,mx=min(mx,max(mi,T-len-c)),c=min(lamx+c+len,T)-mx;
		}
		if (x<mi) ans=mi+c;else
		if (x<=mx) ans=x+c;else
		ans=mx+c;
		if (bz) ans=max(0,ans-(t-r[now-1]));else
		ans=min(T,ans+(t-r[now-1]));
		printf("%d\n",ans);
	}
}