#include<cstdio>
#include<cctype>
#include<algorithm>
using namespace std;
int x,k,q,r[100010],maxn[100010],mini[100010],c[100010];
int rd(){
	int x=0;
	char c;
	do c=getchar();
	while(!isdigit(c));
	do{
		x=(x<<1)+(x<<3)+(c^48);
		c=getchar();
	}while(isdigit(c));
	return x;
}
void InitInput(){
	x=rd(),k=rd();
	for(int i=1;i<=k;i++)
		r[i]=rd();
	maxn[0]=x,r[0]=mini[0]=c[0]=0;
	for(int i=1;i<=k;i++){
		maxn[i]=maxn[i-1];
		mini[i]=mini[i-1];
		if(i&1){
			mini[i]=max(mini[i],min(r[i]-r[i-1]-c[i-1],maxn[i]));
			c[i]=max(c[i-1]+mini[i-1]-r[i]+r[i-1],0)-mini[i];
		}else{
			maxn[i]=min(maxn[i],max(x-r[i]+r[i-1]-c[i-1],mini[i]));
			c[i]=min(c[i-1]+maxn[i-1]+r[i]-r[i-1],x)-maxn[i];
		}
	}
	return;
}
void Ask(){
	q=rd();
	for(int i=1;i<=q;i++){
		int t=rd(),a=rd();
		int pos=upper_bound(r+1,r+k+1,t)-r-1,ans=c[pos];
		if(a<=mini[pos])
			ans+=mini[pos];
		else if(a>=maxn[pos])
			ans+=maxn[pos];
		else
			ans+=a;
		if(pos&1)
			ans=min(x,ans+t-r[pos]);
		else ans=max(0,ans-t+r[pos]);
		printf("%d\n",ans);
	}
}
int main(){
	InitInput();
	Ask();
	return 0;
}
### Prompt

Please formulate a CPP solution to 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<cstdio>
#include<cctype>
#include<algorithm>
using namespace std;
int x,k,q,r[100010],maxn[100010],mini[100010],c[100010];
int rd(){
	int x=0;
	char c;
	do c=getchar();
	while(!isdigit(c));
	do{
		x=(x<<1)+(x<<3)+(c^48);
		c=getchar();
	}while(isdigit(c));
	return x;
}
void InitInput(){
	x=rd(),k=rd();
	for(int i=1;i<=k;i++)
		r[i]=rd();
	maxn[0]=x,r[0]=mini[0]=c[0]=0;
	for(int i=1;i<=k;i++){
		maxn[i]=maxn[i-1];
		mini[i]=mini[i-1];
		if(i&1){
			mini[i]=max(mini[i],min(r[i]-r[i-1]-c[i-1],maxn[i]));
			c[i]=max(c[i-1]+mini[i-1]-r[i]+r[i-1],0)-mini[i];
		}else{
			maxn[i]=min(maxn[i],max(x-r[i]+r[i-1]-c[i-1],mini[i]));
			c[i]=min(c[i-1]+maxn[i-1]+r[i]-r[i-1],x)-maxn[i];
		}
	}
	return;
}
void Ask(){
	q=rd();
	for(int i=1;i<=q;i++){
		int t=rd(),a=rd();
		int pos=upper_bound(r+1,r+k+1,t)-r-1,ans=c[pos];
		if(a<=mini[pos])
			ans+=mini[pos];
		else if(a>=maxn[pos])
			ans+=maxn[pos];
		else
			ans+=a;
		if(pos&1)
			ans=min(x,ans+t-r[pos]);
		else ans=max(0,ans-t+r[pos]);
		printf("%d\n",ans);
	}
}
int main(){
	InitInput();
	Ask();
	return 0;
}
```