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 <cstdio>
#include <iostream>
using namespace std;
const int Q=1<<17;
int s[Q],n,maxn;
int Ch(int x,int l,int r)
{return max(l,min(r,x));}
int main()
{
	scanf("%d%d",&maxn,&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&s[i]);
	int p=1,l=0,r=maxn;
	int qaq,t,a,ty=-1,del=0;
	for(scanf("%d",&qaq);qaq;--qaq){
		scanf("%d%d",&t,&a);
		while(p<=n&&s[p]<=t){
			int val=ty*(s[p]-s[p-1]);
			l=Ch(l+val,0,maxn);
			r=Ch(r+val,0,maxn);
			del+=val;
			ty=-ty;
			++p;
		}
		printf("%d\n",Ch(Ch(a+del,l,r)+ty*(t-s[p-1]),0,maxn));
	}
	return 0;
}