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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int X,k,r[100010],Q;
struct node{int t,a,id;
}q[100010];int ans[100010];
bool cmp(node x,node y) {return x.t<y.t;}
int pls(int x,int y){return max(0,min(X,x+y));}
int main()
{
    scanf("%d%d",&X,&k);
    for (int i=1;i<=k;i++) scanf("%d",&r[i]);
    scanf("%d",&Q);
    for (int i=1;i<=Q;i++) 
    {
        scanf("%d%d",&q[i].t,&q[i].a);q[i].id=i;
    }
    sort(q+1,q+Q+1,cmp);
    int topr=0,topq=1,sum=0,down=0,up=X,add=0,js=-1;
    int cnt;
    while (topq<=Q)
    {
        if (r[topr+1]<q[topq].t&&topr<k)
        {
            topr++;
            add=js*(r[topr]-r[topr-1]);
            down=pls(down,add);up=pls(up,add);
            sum+=add;js=-js;
        } else
        {
            cnt=q[topq].t-r[topr];
            ans[q[topq].id]=pls(max(down,min(up,q[topq].a+sum)),js*cnt);
            topq++;
        }
    }
    for (int i=1;i<=Q;i++) printf("%d\n",ans[i]);
    
}