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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int R[N],ans[N];
struct P{ int t,a,id; inline bool operator < (const P &b) const { return t<b.t; } }p[N];
int main()
{
	int s=gi(),n=gi(),m,i,t,k,l=0,r=s,d,sum=0;
	for (i=1;i<=n;i++) R[i]=gi();
	for (i=1,m=gi();i<=m;i++) p[i].t=gi(),p[i].a=gi(),p[i].id=i;
	sort(p+1,p+1+m);
	for (i=k=1,d=-1;i<=m;i++) {
		for (;k<=n&&R[k]<=p[i].t;k++,d*=-1) {
			l=max(0,l+d*(R[k]-R[k-1]));l=min(l,s);
			r=max(0,r+d*(R[k]-R[k-1]));r=min(r,s);
			sum+=d*(R[k]-R[k-1]);
		}
		t=max(p[i].a+sum,l);t=min(t,r);
		t=max(0,t+d*(p[i].t-R[k-1]));t=min(t,s);
		ans[p[i].id]=t;
	}
	for (i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}
