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 std::max;
using std::min;
int x,n,Q,a,t,r[100005];
int main(){
	scanf("%d",&x);
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&r[i]);
	int lx=0,rx=x,ly=0,ry=x;
	scanf("%d",&Q);
	scanf("%d%d",&t,&a);
	int now=-1;
	for (int i=1;i<=n;i++){
		while (t<=r[i] && Q){
			t=t-r[i-1];
			if (a<=lx) printf("%d\n",min(x,max(ly+t*now,0)));
			else if (a>=rx) printf("%d\n",min(x,max(ry+t*now,0)));
			else printf("%d\n",min(x,max(ly+(a-lx)+t*now,0)));
			Q--;
			if (Q==0) break;
			scanf("%d%d",&t,&a);
		}
		int ti=r[i]-r[i-1]; 
		if (now==-1){
			if (ly-ti<0) lx+=ti-ly;
			if (lx>x){
				lx=x,rx=0;
				ry=0;
			}
			ly=std::max(0,ly-ti);
			ry=std::max(0,ry-ti);
		}else{
			if (ry+ti>x) rx-=(ry+ti-x);
			if (rx<0){
				lx=x,rx=0;
				ly=x;
			}
			ly=std::min(x,ly+ti);
			ry=std::min(x,ry+ti);
		}
		now=now*(-1);
	}
	while (Q){
			t=t-r[n];
			if (a<=lx) printf("%d\n",min(x,max(ly+t*now,0)));
			else if (a>=rx) printf("%d\n",min(x,max(ry+t*now,0)));
			else printf("%d\n",min(x,max(ly+(a-lx)+t*now,0)));
			Q--;
			if (Q==0) break;
			scanf("%d%d",&t,&a);
		}
}