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>
using namespace std;
#define LL long long
#define REP(a,b,c) for(int a=b;a<=c;++a)
#define RREP(a,b,c) for(int a=b;a>=c;--a)

struct node {
	int r,op,a;
}x[300000];
int ans[300000];
int n,m,tim,q,p;
int l,r,y;
int a,b,c,d;

inline bool cmp(node a,node b) {return a.r<b.r;}

int main() {
	scanf("%d%d",&m,&n); y=0,l=0,r=m; d=-1;
	REP(i,1,n) {scanf("%d",&a); x[++p]=(node){a,0,0};}
	scanf("%d",&q); REP(i,1,q) {scanf("%d%d",&a,&b); x[++p]=(node){a,i,b};}
	sort(x+1,x+p+1,cmp);
	REP(i,1,p) {
		if(x[i].op) {
			c=d*(x[i].r-tim);
			if(x[i].a<=l) c+=y;
			else if(x[i].a>=r) c+=y+(r-l);
			else c+=y+(x[i].a-l);
			if(c<0) c=0; if(c>m) c=m; ans[x[i].op]=c;
		} else {
			y+=d*(x[i].r-tim); d=-d; tim=x[i].r;
			if(y<0) {
				if(y+(r-l)<=0) y=l=r=0;
				else {l-=y,y=0;}
			}
			if(y+(r-l)>m) {
				if(y>m) y=l=r=m;
				else r=m+l-y;
			}
		}
	} REP(i,1,q) printf("%d\n",ans[i]); return 0;
}