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;
#define MN 100005
int t,r[MN],a,v[MN],x,n,q;
int main(){
	cin>>x>>n;
	for(int i=1;i<=n;++i)scanf("%d",r+i);
	for(int i=1;i<=n;++i)v[i]=r[i]-r[i-1];
	cin>>q;int tl=0,h=x,l=0,hv=x,lv=0;r[n+1]=v[n+1]=1000000008;
	while(q--){
		scanf("%d%d",&t,&a);
		while(r[tl+1]<=t){
			if((++tl)&1){
				if(lv<v[tl])l+=v[tl]-lv;
				hv=max(0,hv-v[tl]);
				lv=max(0,lv-v[tl]);
			}else{
				if(hv+v[tl]>x)h-=v[tl]+hv-x;
				hv=min(x,hv+v[tl]);
				lv=min(x,lv+v[tl]);
			}
			h=max(0,h);l=min(l,x);
		}t-=r[tl];
		if(l>=a)a=lv;else if(h<=a)a=hv;else a+=lv-l;
		if(tl&1)printf("%d\n",min(x,a+t));else printf("%d\n",max(0,a-t));
	}
	return 0;
}