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<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>

using namespace std;
typedef long long LL;

const int MX=100011;

LL x,k;int q;
LL r[MX];
struct Func{
	LL k,b,L,R;//A=min(X,max(0,kt+b+a)),in[L,R]
	Func(int _k=0,int _b=0,int _L=0,int _R=0):k(_k),b(_b),L(_L),R(_R){}
	inline LL V(LL a,LL t){return min(x,max(0LL,k*t+b+min(R,max(L,a))));}
}f[MX];

int main(){
	scanf("%lld%lld",&x,&k);
	for(int i=1;i<=k;i++)scanf("%lld",&r[i]);
	int K=-1;
	Func cur(K,0,0,x);
	f[0]=cur;
	for(int i=1;i<=k;i++){
		LL vl=f[i-1].V(f[i-1].L,r[i]),vr=f[i-1].V(f[i-1].R,r[i]);
		K=-K;
		if(vl>0&&vr<x){
			f[i]=cur=Func(K,vl-f[i-1].L-K*r[i],f[i-1].L,f[i-1].R);
		}else if(vr<x&&vr>0){
			f[i]=cur=Func(K,vr-f[i-1].R-K*r[i],f[i-1].R-vr,f[i-1].R);
		}else if(vl>0&&vl<x){
			f[i]=cur=Func(K,vl-f[i-1].L-K*r[i],f[i-1].L,f[i-1].L+x-vl);
		}else{//(vl==vr)
			f[i]=cur=Func(K,vl-K*r[i],0,0);
		}
	}
	scanf("%d",&q);
	for(int i=1;i<=q;i++){
		LL t,a;scanf("%lld%lld",&t,&a);
		printf("%lld\n",f[upper_bound(r,r+k+1,t)-r-1].V(a,t));
	}
	return 0;
}
