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;
int x,k,q,t,a,mx[100005],mn[100005],dt[100005],rv[100005],ans;
int main()
{
	scanf("%d%d",&x,&k);mx[0]=x;mn[0]=0;dt[0]=0;
	for(int i=1;i<=k;i++) scanf("%d",&rv[i]);
	for(int i=1;i<=k;i++)
	{
		mx[i]=mx[i-1];mn[i]=mn[i-1];int tmp=rv[i]-rv[i-1];
		if(i&1) mn[i]=max(mn[i],min(mx[i],tmp-dt[i-1])),dt[i]=max(mn[i-1]+dt[i-1]-tmp,0)-mn[i];
		else mx[i]=min(mx[i],max(mn[i],x-tmp-dt[i-1])),dt[i]=min(mx[i-1]+dt[i-1]+tmp,x)-mx[i];
	}
	for(scanf("%d",&q);q--;printf("%d\n",ans))
	{
		scanf("%d%d",&t,&a);int pos=upper_bound(rv+1,rv+k+1,t)-rv-1;
		if(a<=mn[pos]) a=mn[pos];else if(a>=mx[pos]) a=mx[pos];
		ans=a+dt[pos];int tmp=t-rv[pos];
		if(pos&1) ans=min(x,ans+tmp);else ans=max(0,ans-tmp);
	}
}