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<vector>
#include<algorithm>
using namespace std;
int X,K;
vector<pair<int,int> >A;
int main()
{
	cin>>X>>K;
	for(int i=0;i<K;i++)
	{
		int r;cin>>r;A.push_back(make_pair(r,-1));
	}
	int Q;cin>>Q;
	for(int i=0;i<Q;i++)
	{
		int t,a;cin>>t>>a;
		A.push_back(make_pair(t,a));
	}
	sort(A.begin(),A.end());
	int pre=0,d=1;
	int L=0,R=X,AA=0;
	for(pair<int,int>p:A)
	{
		if(p.second<0)
		{
			int t=p.first-pre;
			pre=p.first;
			if(d>0)
			{
				AA-=t;
				if(L+AA<0)
				{
					L=-AA;
				}
				if(L>R)R=L;
			}
			else
			{
				AA+=t;
				if(R+AA>X)
				{
					R=X-AA;
				}
				if(L>R)L=R;
			}
			d*=-1;
		}
		else
		{
			int a=p.second;
			if(a<L)a=L;
			else if(a>R)a=R;
			a+=AA;
			int t=p.first-pre;
			cout<<(d>0?max(0,a-t):min(a+t,X))<<endl;
		}
	}
}
