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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100010;
int n,m,a[maxn];
int calc(int l,int r,int x)
{return max(l,min(r,x));}
int main(){
	int X;
	scanf("%d%d",&X,&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	int L=0,R=X;//初值……习惯 
	int t=0,k=0,s=-1,x=0;
	int time,A;
	scanf("%d",&m);
	for(int i=1;i<=m;i++){
		scanf("%d%d",&time,&A);
		while(k<n&&a[k+1]<=time){
			int dif=s*(a[k+1]-t);
			L=calc(0,X,L+dif);
			R=calc(0,X,R+dif);
			s*=-1;
			x+=dif;
			t=a[k+1];
			k++;
		}//printf("\nL=%d R=%d\n",L,R);
		//printf("\nk=%d\n",k);
		int T=time-t;
		A=calc(L,R,A+x);
		//printf("\nA=%d\n",A);
		A=calc(0,X,A+s*T);
		printf("%d\n",A);
	}
	return 0;
}