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>
int X,k,q;
struct ev{
	int x,y,id;
	bool operator<(const ev&e)const{return x<e.x;}
}e[500007];
int ep=0,as[500007];
int main(){
	scanf("%d%d",&X,&k);
	for(int i=0,a;i<k;++i){
		scanf("%d",&a);
		e[ep++]=(ev){a,0,0};
	}
	scanf("%d",&q);
	for(int i=1,a,b;i<=q;++i){
		scanf("%d%d",&a,&b);
		e[ep++]=(ev){a,b,i};
	}
	std::sort(e,e+ep);
	long long mn=0,mx=X,a=0;
	for(int i=0,sgn=-1,t0=0;i<ep;++i){
		long long v0=(e[i].x-t0)*sgn;
		if(e[i].id){
			long long v=e[i].y+a;
			if(v<mn)v=mn;
			if(v>mx)v=mx;
			v+=v0;
			if(v<0)v=0;
			if(v>X)v=X;
			as[e[i].id]=v;
		}else{
			mn+=v0,mx+=v0,a+=v0;
			if(mn<0) mn=0;
			if(mx<0) mx=0;
			if(mx>X) mx=X;
			if(mn>X) mn=X; 
			t0=e[i].x;
			sgn*=-1;
		}
	}
	for(int i=1;i<=q;++i)printf("%d\n",as[i]);
	return 0;
} 