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;
#define int long long
int x;
vector<int>rev;
int ptr=0;
#undef int
int main(){
	#define int long long
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>x;
	int k;cin>>k;
	rev.push_back(0);
	for(int i=0;i<k;i++){
		int t;cin>>t;
		rev.push_back(t);
	}
	int q;cin>>q;
	int d=0,u=x,dans=0,uans=x;
	for(int i=0;i<q;i++){
		int t,a;cin>>t>>a;
		while(ptr<rev.size()-1&&t>=rev[ptr+1]){
			if(ptr%2==0){
				dans-=(rev[ptr+1]-rev[ptr]);
				if(dans<0){
					d+=(-dans);
					dans=0;
				}
				if(d>u){
					d=u;uans=0;
				}else{
					uans-=(rev[ptr+1]-rev[ptr]);
				}
			}else{
				uans+=(rev[ptr+1]-rev[ptr]);
				if(uans>x){
					u-=(uans-x);
					uans=x;
				}
				if(u<d){
					u=d;dans=x;
				}else{
					dans+=(rev[ptr+1]-rev[ptr]);
				}
			}
			ptr++;
		}
		if(ptr%2==0){
			if(a>=u){
				cout<<max(0LL,uans-(t-rev[ptr]))<<"\n";
			}else if(a<=d){
				cout<<max(0LL,dans-(t-rev[ptr]))<<"\n";
			}else{
				int stans=(a-d)+dans;
				cout<<max(0LL,stans-(t-rev[ptr]))<<"\n";
			}
		}else{
			if(a>=u){
				cout<<min(x,uans+(t-rev[ptr]))<<"\n";
			}else if(a<=d){
				cout<<min(x,dans+(t-rev[ptr]))<<"\n";
			}else{
				int stans=(a-d)+dans;
				cout<<min(x,stans+(t-rev[ptr]))<<"\n";
			}
		}
	}
	return 0;
}