#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+20;
int X,K,Q,solve();
int hehe=solve(),r[maxn],mx[maxn],mn[maxn],c[maxn];
int solve(){
	scanf("%d",&X);
	scanf("%d",&K);
	for(int i=1;i<=K;++i){
		scanf("%d",&r[i]); 
	}
	mx[0]=X;
	for(int i=1;i<=K;++i){
		mx[i]=mx[i-1];mn[i]=mn[i-1];
		if(i&1){
			mn[i]=max(mn[i],min(r[i]-r[i-1]-c[i-1],mx[i]));
			c[i]=max(c[i-1]+mn[i-1]-r[i]+r[i-1],0)-mn[i]; 
		}
		else{
			mx[i]=min(mx[i],max(X-r[i]+r[i-1]-c[i-1],mn[i]));
			c[i]=min(c[i-1]+mx[i-1]+r[i]-r[i-1],X)-mx[i];
		}
	}
	scanf("%d",&Q);
	for(int a,t,pl,ans;Q--;){
		scanf("%d%d",&t,&a);
		ans=c[pl=(upper_bound(r+1,r+K+1,t)-r-1)];
		if(a<=mn[pl])ans+=mn[pl];
		else if(a>=mx[pl])ans+=mx[pl];
		else ans+=a;
		if(pl&1)ans=min(X,ans+t-r[pl]);
		else ans=max(0,ans-t+r[pl]);
		printf("%d\n",ans);     
	}
	return 0;
}
int main(){};
### 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 maxn=1e5+20;
int X,K,Q,solve();
int hehe=solve(),r[maxn],mx[maxn],mn[maxn],c[maxn];
int solve(){
	scanf("%d",&X);
	scanf("%d",&K);
	for(int i=1;i<=K;++i){
		scanf("%d",&r[i]); 
	}
	mx[0]=X;
	for(int i=1;i<=K;++i){
		mx[i]=mx[i-1];mn[i]=mn[i-1];
		if(i&1){
			mn[i]=max(mn[i],min(r[i]-r[i-1]-c[i-1],mx[i]));
			c[i]=max(c[i-1]+mn[i-1]-r[i]+r[i-1],0)-mn[i]; 
		}
		else{
			mx[i]=min(mx[i],max(X-r[i]+r[i-1]-c[i-1],mn[i]));
			c[i]=min(c[i-1]+mx[i-1]+r[i]-r[i-1],X)-mx[i];
		}
	}
	scanf("%d",&Q);
	for(int a,t,pl,ans;Q--;){
		scanf("%d%d",&t,&a);
		ans=c[pl=(upper_bound(r+1,r+K+1,t)-r-1)];
		if(a<=mn[pl])ans+=mn[pl];
		else if(a>=mx[pl])ans+=mx[pl];
		else ans+=a;
		if(pl&1)ans=min(X,ans+t-r[pl]);
		else ans=max(0,ans-t+r[pl]);
		printf("%d\n",ans);     
	}
	return 0;
}
int main(){};
```