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 MAX 100007
int r[MAX];
int a[MAX];
int b[MAX];
int k1[MAX];

int main(){
	int x,k,q;
	cin >> x >> k;
	for(int i=1; i<=k; i++)cin >> r[i];
	r[k+1]=INT_MAX;
	a[0]=0;
	b[0]=x;
	k1[0]=1;
	// printf("%d %d %d\n",a[0],b[0],k1[0]);
	for(int i=1; i<=k; i++){
		int dt=r[i]-r[i-1];
		if(i&1){
			a[i]=max(a[i-1]-dt,0);
			b[i]=max(b[i-1]-dt,0);
			if(a[i]==b[i])k1[i]=(x+1);
			else k1[i]=min(k1[i-1]+max(dt-a[i-1],0),x);
		}
		else{
			a[i]=min(a[i-1]+dt,x);
			b[i]=min(b[i-1]+dt,x);
			if(a[i]==b[i])k1[i]=(x+1);
			else k1[i]=max(k1[i-1]-max(dt-x+a[i-1]+1,0),0);
		}
		// printf("%d %d %d\n",a[i],b[i],k1[i]);
	}
	cin >> q; 
	for(int i=1; i<=q; i++){
		int tq, aq;
		cin >> tq >> aq;
		if(tq==0){cout << aq << endl; continue;}
		int p=lower_bound(r,r+k+2,tq)-r-1;
		int dt=tq-r[p];
		int	res=min(a[p]+max(aq-k1[p]+1,0),b[p]);
		// cout << dt << " " << res << " ";
		res= (p&1) ? min(res+dt,x) :max(res-dt,0);
		// cout << p << " " << res << endl;
		cout << res << endl;
	}
}