#include<bits/stdc++.h>
using namespace std;

const int N=100100;
int a[N],X,n,q,sum;
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>X>>n;
	for(int i=1;i<=n;++i)cin>>a[i];
	int L=0,R=X,tag=-1,i=1; 
	for(cin>>q;q --> 0;){
		int x,y;cin>>x>>y;
		for(;i<=n&&a[i]<x;++i){
			int tmp=tag*(a[i]-a[i-1]);
			L+=tmp;L=max(0,min(X,L));
			R+=tmp;R=max(0,min(X,R));
			sum+=tmp;tag=-tag;
		}
		int tmp=max(L,min(R,y+sum))+tag*(x-a[i-1]);
		cout<<max(0,min(X,tmp))<<'\n';
	}
	return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int N=100100;
int a[N],X,n,q,sum;
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>X>>n;
	for(int i=1;i<=n;++i)cin>>a[i];
	int L=0,R=X,tag=-1,i=1; 
	for(cin>>q;q --> 0;){
		int x,y;cin>>x>>y;
		for(;i<=n&&a[i]<x;++i){
			int tmp=tag*(a[i]-a[i-1]);
			L+=tmp;L=max(0,min(X,L));
			R+=tmp;R=max(0,min(X,R));
			sum+=tmp;tag=-tag;
		}
		int tmp=max(L,min(R,y+sum))+tag*(x-a[i-1]);
		cout<<max(0,min(X,tmp))<<'\n';
	}
	return 0;
}

```