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>
const int MaxK=100010,MaxQ=100010;
int X,K,r[MaxK],Q,t[MaxQ],a[MaxQ];
int L,R,b,T;
bool dir;
int min(int a,int b){return a<b?a:b;}
int max(int a,int b){return a>b?a:b;}
void mov(int t){
	if(dir){
		b+=min(t-T,X-b-L);
		R=min(R,X-b);
	}
	else{
		b-=min(t-T,b+R);
		L=max(L,-b);
	}
	T=t;
}
int main(){
	scanf("%d%d",&X,&K);
	for(int i=0;i<K;i++)scanf("%d",r+i);
	scanf("%d",&Q);
	for(int i=0;i<Q;i++)scanf("%d%d",t+i,a+i);
	R=X;
	for(int i=0,j=0;i<K||j<Q;){
		if(j==Q||i<K&&r[i]<t[j]){
			mov(r[i++]);
			dir^=1;
		}
		else{
			mov(t[j]);
			printf("%d\n",b+min(max(a[j++],L),R));
		}
	}
}