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>
#define L long long
#define pb push_back
#define pi pair<int,int>
#define pii pair<pair<int,int>,int>
#define aa first
#define bb second
#define mp make_pair
#define mpp(a,b,c) mp(mp(a,b),c)
using namespace std;
int n,m,q,x[100010],a,b,c,t;
inline void orz(int t,int k)
{
	if(k==1)
	  {
	   c+=t;
	   if(b+c>m)
	     b=m-c;
	   if(a>b)
	     a=b=0,c=m;
	  }
	else
	  {
	   c-=t;
	   if(a+c<0)
	     a=-c;
	   if(a>b)
		 a=b=c=0;
	  }
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k,l;
	scanf("%d%d",&m,&n);
	for(i=1;i<=n;i++)
	  scanf("%d",&x[i]);
	b=m;
	k=1;
	l=-1;
	scanf("%d",&q);
	while(q--)
	  {
	   scanf("%d%d",&i,&j);
	   while(k<=n && i>=x[k])
	     {
		  orz(x[k]-t,l);
		  t=x[k];
		  l*=-1;
		  k++;
		 }
	   orz(i-t,l);
	   t=i;
	   if(j<=a)
	     printf("%d\n",a+c);
	   else if(j<=b)
	     printf("%d\n",j+c);
	   else
	     printf("%d\n",b+c);
	  }
	return 0;
}
