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>
using namespace std;
#define N 1111116

int n,m,x,mx,mn,k,b,a[N],ans[N];
pair<int,int> q[N];

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	x=read();n=read();
	for (int i=1;i<=n;i++)
		q[i]=make_pair(read(),0);
	m=read();
	for (int i=1;i<=m;i++){
		q[n+i]=make_pair(read(),i);
		a[i]=read();
	}
	sort(q+1,q+n+m+1);
	mn=0;mx=x;k=-1;b=0;
	for (int i=1;i<=n+m;i++){
		mn+=k*(q[i].first-q[i-1].first);
		mn=min(max(mn,0),x);
		mx+=k*(q[i].first-q[i-1].first);
		mx=min(max(mx,0),x);
		b+=k*(q[i].first-q[i-1].first);
		if (!q[i].second) k=-k;
		else ans[q[i].second]=min(max(a[q[i].second]+b,mn),mx);
	}
	for (int i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}