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<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int M=1e5+7;
int read(){
	int ans=0,f=1,c=getchar();
	while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
	while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
	return ans*f;
}
int ans[M],T[M];
struct pos{int T,s;}e[M];
int x,k=1,n,m,sgn=-1;
void calc(LL &k){if(k<0) k=0; if(k>x) k=x;}
int main(){
	x=read();
	n=read(); for(int i=1;i<=n;i++) T[i]=read();
	m=read(); for(int i=1;i<=m;i++) e[i].T=read(),e[i].s=read();
	LL h=0,mx=x,mn=0;
	for(int i=1;i<=m;i++){
		while(k<=n&&T[k]<=e[i].T){
			LL v=(T[k]-T[k-1])*sgn;
			mx+=v; mn+=v; h+=v;
			calc(mx); calc(mn);
			k++;sgn*=-1; 
		}//printf("[%lld %lld %lld]\n",mx,mn,h);
		LL now=e[i].s+h,nowh=(e[i].T-T[k-1])*sgn; //printf("[%lld %lld]\n",now,nowh);
		if(now<mn) now=mn;
		if(now>mx) now=mx;
		now+=nowh; calc(now);
		printf("%lld\n",now);
	}
	return 0;
}