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>
using namespace std;
const int N=1e5+10;
int sm,n,t,q,a,r[N],v[N];
int main(){
	scanf("%d%d",&sm,&n);
	for (int i=1;i<=n;++i) scanf("%d",r+i);
	for (int i=n;i;--i) v[i]=r[i]-r[i-1];
	int low=0,lh=0;
	int up=sm,uh=sm;
	int pos=0;r[n+1]=2e9;
	scanf("%d",&q);
	while (q--){
		scanf("%d%d",&t,&a);
		while (r[pos+1]<=t){
			++pos;
			if (pos&1){//ÏòÏÂv[pos] 
				if (lh<v[pos]) low+=v[pos]-lh;
				lh=max(lh-v[pos],0);
				uh=max(uh-v[pos],0);
			}else{//ÏòÉÏ 
				if (uh+v[pos]>sm) up-=uh+v[pos]-sm;
				lh=min(lh+v[pos],sm);
				uh=min(uh+v[pos],sm);
			}
			low=min(low,sm);
			up=max(up,0);
		}
		t-=r[pos];
		if (a<=low) a=lh;
		else if (a>=up) a=uh;
		else a=lh+a-low;
		if (pos&1) printf("%d\n",min(a+t,sm));else printf("%d\n",max(a-t,0));
	}
	return 0;
}