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 <string>
#include <vector>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <map>
#include <list>
#include <set>
#include <numeric>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <climits>
#include <cfloat>
#include <ctime>
#include <complex>
#include <cassert>
#include <array>
#include <bitset> 
#include <unordered_map>
#include <random>

using namespace std;
typedef long long LL;
typedef pair<int,int> P;

const int L=1e5+1;
int X;
int K;
int r[L];
int b[L];
int u[L];
int l[L];
int main() {
	scanf("%d%d",&X,&K);
	for(int i=1;i<=K;i++){
		scanf("%d",r+i);
	}
	u[0]=X;
	for(int i=1;i<=K;i++){
		int d;
		if(i%2){
			d=-r[i]+r[i-1];
		}
		else{
			d=r[i]-r[i-1];
		}
		b[i]=b[i-1]+d;
		u[i]=u[i-1]+d;
		l[i]=l[i-1]+d;
		u[i]=max(0,min(X,u[i]));
		l[i]=max(0,min(X,l[i]));
	}
	int Q;
	scanf("%d",&Q);
	for(int i=0;i<Q;i++){
		int t,a;
		scanf("%d%d",&t,&a);
		int p=upper_bound(r,r+K+1,t)-r;
		p--;
		int s=a+b[p];
		if(s<l[p]){
			s=l[p];
		}
		else if(u[p]<s){
			s=u[p];
		}
		if(p%2==0){
			s-=t-r[p];
		}
		else{
			s+=t-r[p];
		}
		s=max(0,min(X,s));
		printf("%d\n",s);
	}




	return 0;
}

