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 <cstdio>
#include <cstring>
#include <cassert>
#include <cctype>
using namespace std;
typedef long long lint;
#define cout cerr
#define ni (next_num<int>())
template<class T>inline T next_num(){
	T i=0;char c;
	while(!isdigit(c=getchar())&&c!='-');
	bool flag=c=='-';
	flag?c=getchar():0;
	while(i=i*10-'0'+c,isdigit(c=getchar()));
	return flag?-i:i;
}
template<class T1,class T2>inline void apmax(T1 &a,const T2 &b){if(a<b)a=b;}
template<class T1,class T2>inline void apmin(T1 &a,const T2 &b){if(b<a)a=b;}
const int N=100010;
int lim;
int r[N];
int lb,hb;
int tim=0;
lint mn=0,cur=0,mx=0;
inline int go(int x,int d,int t){
	return max(0ll,min((lint)lim,(lint)d*t+x));
}
inline void workto(int d,int t){
	int del=t-tim;
	tim=t;
	lb=go(lb,d,del),hb=go(hb,d,del);
	cur+=(lint)d*del;
	apmin(mn,cur);
	apmax(mx,cur);
}
inline int ask(int x){
	if(x<=-mn)return lb;
	else if(x>=lim-mx)return hb;
	else return x+cur;
}
int main(){
	lb=0,lim=hb=ni;
	int n=ni;
	r[0]=0;
	for(int i=1;i<=n;i++){
		r[i]=ni;
	}
	int d=-1;
	for(int i=1,tot=ni;tot--;){
		int tim=ni;
		for(;i<=n&&r[i]<=tim;workto(d,r[i]),d=-d,i++);
		workto(d,tim);
		printf("%d\n",ask(ni));
	}
	return 0;
}