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<cstdio>
#include<algorithm>

using namespace std;

int X,K,Q,maxn[100001],minn[100001],r[1000001],f[1000001],t,a,ans;

int main(){
	scanf("%d%d",&X,&K);maxn[0]=X;for(int i=1;i<=K;++i)scanf("%d",r+i);
	for(int i=1;i<=K;++i){
		int t=r[i]-r[i-1];maxn[i]=maxn[i-1],minn[i]=minn[i-1];
		if(i&1)minn[i]=max(minn[i],min(maxn[i],t-f[i-1])),f[i]=max(minn[i-1]+f[i-1]-t,0)-minn[i];
		else maxn[i]=min(maxn[i],max(minn[i],X-t-f[i-1])),f[i]=min(maxn[i-1]+f[i-1]+t,X)-maxn[i];
	}
	scanf("%d",&Q);
	for(int i=1;i<=Q;++i){
		scanf("%d%d",&t,&a);
		int j=upper_bound(r+1,r+K+1,t)-r-1,p=t-r[j];
		a=max(minn[j],a),a=min(maxn[j],a);
		ans=a+f[j];
		if(j&1)ans=min(X,ans+p);else ans=max(0,ans-p);
		printf("%d\n",ans);
	}
}