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;

const int N=100005;

int m,n,Q;
struct func{
	int a,b,w,t,k;
	int g(int x){
		if(x<=a)return w;
		if(x<=b)return w+x-a;
		return w+b-a;
	}
	void expand(int dt){
		t+=dt;w+=dt*k;k=-k;
		if(w+b-a<=0)a=b=m,w=0;else if(w<0)a-=w,w=0;
		if(w>=m)a=b=m,w=m;else if(w+b-a>m)b-=w+b-a-m;
	}
}f[N];
bool operator<(const int&a,const func&b){return a<b.t;}

int main(){
	scanf("%d%d",&m,&n);
	f[0]=(func){0,m,0,0,-1};
	for(int i=1,t;i<=n;i++)f[i]=f[i-1],scanf("%d",&t),f[i].expand(t-f[i].t);
	scanf("%d",&Q);
	for(int t,a,p,r;Q--;){
		scanf("%d%d",&t,&a);
		p=upper_bound(f,f+n+1,t)-f-1;
		r=f[p].g(a)+(t-f[p].t)*f[p].k;
		if(r<0)r=0;if(r>m)r=m;
		printf("%d\n",r);
	}
	return 0;
}