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<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 220000
using namespace std;
int a[N],b[N],sum[N],l[N],r[N],A,B,ans0[N],ans1[N];
int q,n,x;
int main(){
	scanf("%d",&x);
	scanf("%d",&n);
	n++;
	for(int i=1;i<=n;i++){
		if(i<n)scanf("%d",&a[i]);
		else a[i]=1200000000;
		b[i]=a[i]-a[i-1];
		b[i]=b[i]*((i&1)?-1:1);
		sum[i]=sum[i-1]+b[i];
	}
	l[0]=0,r[0]=x;
	for(int i=1;i<=n;i++){
		l[i]=l[i-1];
		r[i]=r[i-1];
		l[i]=max(l[i],-sum[i]);
		r[i]=min(r[i],x-sum[i]);
	}
	//for(int i=1;i<=n;i++)printf("%d %d\n",l[i],r[i]);
	A=0,B=x;
	for(int i=1;i<=n;i++){
		A=A+b[i];
		B=B+b[i];
		A=max(A,0);A=min(A,x);
		B=max(B,0);B=min(B,x);
		ans0[i]=A;
		ans1[i]=B;
	}
//	for(int i=1;i<=n;i++)printf("%d ",b[i]);printf("\n");
//	for(int i=1;i<=n;i++)printf("%d ",ans1[i]);printf("\n");
	scanf("%d",&q);
	for(int i=1;i<=q;i++){
		int X,y;
		scanf("%d%d",&X,&y);
		int p=lower_bound(a+1,a+1+n,X)-a;
		if(a[p]>X)p--;
	//	printf("%d %d %d\n",X,y,p);
		int ans=0;
		if(y<=l[p]){
			ans=ans0[p];
		}
		if(y>=r[p]){
			ans=ans1[p];
		}
		if(y>=l[p]&&y<=r[p]){
			ans=y+sum[p];
		}
		if(p&1)ans+=X-a[p];
		else ans-=X-a[p];
		ans=max(ans,0);
		ans=min(ans,x);
		printf("%d\n",ans);
	}
}