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<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int x,k,r[100050],s[100050],o[100050],w[100050],q,t,a;
int main(){
	scanf("%d%d",&x,&k);
	int i,j,v=-1;
	r[0]=o[0]=w[0]=0;
	s[0]=x;
	for(i=1;i<=k;i++){
		scanf("%d",&r[i]);
		int d=(r[i]-r[i-1])*v;
		w[i]=min(max(w[i-1]+d,0),x);
		s[i]=max(min(s[i-1]+d,x),0);
		o[i]=min(o[i-1]-min(0,w[i-1]+d),x);
		v*=-1;
	}
	i=0;	
	scanf("%d",&q);
	while(q--){
		scanf("%d%d",&t,&a);
		if(t>=r[k]) i=k;
		else{
			int n=k;
			while(i<n-1){
				int c=i+(n-i)/2;
				if(r[c]<t) i=c;
				else n=c;
			}
		}
		int ans=min(w[i]+max(a-o[i],0),s[i])+(t-r[i])*((i%2)*2-1);
		printf("%d\n",min(max(ans,0),x));
	}
	return 0;
}