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 mo 1000000007
#define pi 3.1415926535898
#define eps 1e-9 
using namespace std;
long long read(){
    long long xx=0,flagg=1;
    char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')
        ch=getchar();
    if(ch=='-'){
        flagg=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        xx=xx*10+ch-'0';
        ch=getchar();
    }
    return xx*flagg;
}
void pus(long long xx,long long flagg){
    if(xx<0){
        putchar('-');
        xx=-xx;
    }
    if(xx>=10)
        pus(xx/10,0);
    putchar(xx%10+'0');
    if(flagg==1)
        putchar(' ');
    if(flagg==2)
        putchar('\n');
    return;
}
int X,n,m,i,r[100005],L,R,ti,a,flag,x,sum;
int cz(int u,int v,int w){
	return max(u,min(v,w));
}
int main(){
   	X=read();n=read();
    for(i=1;i<=n;i++)
    	r[i]=read();
    m=read();
   	R=X;i=0;flag=-1;
    while(m--){
        ti=read();a=read();
        while(i<n&&r[i+1]<=ti){
            x=flag*(r[i+1]-r[i]);
            L=cz(0,X,L+x);
            R=cz(0,X,R+x);
            flag=-flag;
            sum+=x;
            i++;
        }
        pus(cz(0,X,cz(L,R,a+sum)+flag*(ti-r[i])),2);
    }
    return 0;
}