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>
#include<cstdlib>
#include<algorithm>
#define N 100010
#define ll long long
using namespace std;
ll num[N],ans[N];
struct ques{
	int a,t,id;
	bool operator <(const ques b)const{return t<b.t;}
}q[N];
int main()
{
	ll x;
	int n;
	scanf("%lld%d",&x,&n);
	for(int i=1;i<=n;i++) scanf("%lld",&num[i]);
	int m;
	scanf("%d",&m);
	for(int i=1;i<=m;i++) scanf("%d%d",&q[i].t,&q[i].a),q[i].id=i;
	sort(q+1,q+m+1);
	ll u=1,r=0,up=x,down=0,opt=-1;
	ll sum=0;
	while(u<=m)
	{
		if(num[r+1]<q[u].t && r<n)
        {
        	ll res=0;
            r++;
            res=opt*(num[r]-num[r-1]);
            down=max(0ll,min(x,down+res));
			up=max(0ll,min(x,up+res));
            sum+=res;
			opt=-opt;
        }
		else
		{
			ll cnt=q[u].t-num[r];
			ll res=max(down,min(up,q[u].a+sum));
			ans[q[u].id]=max(0ll,min(x,res+opt*cnt));
			u++;
        }
    }
	for(int i=1;i<=m;i++) printf("%lld\n",ans[i]);
	return 0;
}
