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<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=100000+10;
struct dong{
	int t,a;
} ask[maxn];
int p[maxn],ans[maxn];
int i,j,k,l,t,n,m,f,a,b,c,d,e,X;
bool czy;
int get(int x){
	if (czy) return e;
	if (x<=a) return b;
	else if (x>=c) return d;
	else return (x-a)+b;
}
void change(int f,int t){
	if (czy){
		e+=f*t;
		e=max(0,min(X,e));
		return;
	}
	if (f==-1){
		if (d-t<=0){
			czy=1;
			e=0;
			return;
		}
		if (t>=b){
			a+=t-b;
			b=0;
			d-=t;
		}
		else{
			b-=t;
			d-=t;
		}
	}
	else{
		if (b+t>=X){
			czy=1;
			e=X;
			return;
		}
		if (d+t>=X){
			b+=t;
			c-=d+t-X;
			d=X;
		}
		else{
			b+=t;
			d+=t;
		}
	}
}
int main(){
	scanf("%d",&X);
	scanf("%d",&n);
	fo(i,1,n) scanf("%d",&p[i]);
	scanf("%d",&m);
	fo(i,1,m) scanf("%d%d",&ask[i].t,&ask[i].a);
	j=0;
	a=0;b=0;c=X;d=X;
	f=-1;
	fo(i,1,n){
		while (j<m&&ask[j+1].t<p[i]){
			j++;
			k=ask[j].a;
			ans[j]=max(0,min(X,get(k)+f*(ask[j].t-p[i-1])));
		}
		change(f,p[i]-p[i-1]);
		f*=-1;
	}
	while (j<m){
		k=ask[++j].a;
		ans[j]=max(0,min(X,get(k)+f*(ask[j].t-p[n])));
	}
	fo(i,1,m) printf("%d\n",ans[i]);
}