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<bits/stdc++.h>
using namespace std;
const int N = 100005;
int X,K,Q,r[N],t[N],x[N],la,lb,a,b,c,t1,t2,add,tmp,diff,f;
int main() {
//	freopen("data.in","r",stdin);
	cin>>X;
	cin>>K;
	for(int i=1;i<=K;i++)
		cin>>r[i];
	cin>>Q;
	for(int i=1;i<=Q;i++)
		cin>>t[i]>>x[i];
	t1=0,t2=1;
	a=0,b=X,f=-1;
	while(t2<=Q) {
		if(t[t2]>r[t1+1]&&t1<K) {
			t1++;
			add=f*(r[t1]-r[t1-1]);
//			b=max(0ll,max(0ll,min(X,X-f*add))-c);
//			a=min(X,min(X,max(0ll,0ll-f*add))-c);
			b+=add,a+=add;
			b=max(0,min(X,b)),a=max(0,min(X,a));
			c+=add;
			f*=-1;
		}
		else {
			diff=t[t2]-r[t1];
			tmp=max(a,min(b,x[t2]+c));
			tmp+=diff*f;
			tmp=max(0,min(X,tmp));
			cout<<tmp<<endl;
			t2++;
		}
	}
	return 0;
}