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<cstring>
#include<algorithm>
#define N 100010
using namespace std;

int n, m, t[N], w[N], a[N], l, r, lx, rx, sum, x, y, now, q;

int main(){
	scanf("%d%d", &m, &n);
	for(int i=1; i<=n; i++)scanf("%d", &t[i]);
	scanf("%d", &q);
	for(int i=1; i<=q; i++)scanf("%d%d", &w[i], &a[i]);
	t[0]=0; t[n+1]=w[q]+1; now=0;
	l=0; lx=0; r=m; rx=m; sum=0;
	while(now<q&&w[now+1]<t[1]){now++; printf("%d\n", max(0, a[now]-w[now]));}
	for(int i=1; i<=n; i++){
		x=t[i]-t[i-1];
		if(i&1){
			if(l+1<=r-1&&l+1+sum-x<=0)l=min(m, x-sum);
			if(l>=r)l=m; lx=max(0, lx-x);
			if(l==m)r=1; rx=max(0, rx-x);
			sum-=x;
		}else{
			if(l+1<=r-1&&r-1+sum+x>=m)r=max(0, m-x-sum);
			if(r<=l)r=0; rx=min(m, rx+x);
			if(!r)l=m; lx=min(m, lx+x);
			sum+=x;
		}
		while(now<q&&w[now+1]<t[i+1]){
			now++;
			if(a[now]<=l)y=lx;
			if(l<a[now]&&a[now]<r)y=a[now]+sum;
			if(a[now]>=r)y=rx;
			if(i&1)printf("%d\n", min(m, y+w[now]-t[i]));
			else printf("%d\n", max(0, y-(w[now]-t[i])));
		}
	}
	return 0;
}