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>
#define ll long long
#define N 100009
using namespace std;

int n,a[N],b[N]; ll m;
int main(){
	scanf("%lld%d",&m,&n);
	int i;
	for (i=1; i<=n; i++){
		scanf("%d",&a[i]); b[i]=a[i]-a[i-1];
		if (i&1) b[i]=-b[i];
	}
	int cas; scanf("%d",&cas);
	ll l=0,r=m,x=0,y=m,tg=0,u,v; i=1;
	while (cas--){
		scanf("%lld%lld",&u,&v);
		for (; i<=n && u>=a[i]; i++){
			tg+=b[i]; x+=b[i]; y+=b[i];
			x=min(m,max(0ll,x));
			y=min(m,max(0ll,y));
			r=min(r,m-tg); l=max(l,-tg);
		}
		if (l<=v && v<=r) v+=tg;
		else v=v<l?x:y;
		if (i&1) v-=u-a[i-1]; else v+=u-a[i-1];
		printf("%lld\n",min(m,max(0ll,v)));
	}
	return 0;
}
