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>
using namespace std;
typedef long long ll;
const int max_n=200005;
struct Node{
    int t,a,id;
    bool operator < (Node p)const {
        return t<p.t;
    }
}qry[max_n];
int X,k,q;
bool down;
int a,b,c;
void update(int v){
    int u=b-a+c;
    if(u+v>X){
        b-=u+v-X;
        c+=v;
        b=max(b,a),c=min(c,X);
    }
    else if(c+v<0){
        a-=c+v;
        c+=v;
        a=min(a,b),c=max(c,0);
    }
    else c+=v;
}
int query(int x){
    //cout<<x<<' '<<a<<' '<<b<<' '<<c<<endl;
    if(x<a)return c;
    else if(x>=a&&x<=b)return x-a+c;
    else return b-a+c;
}
int ans[max_n];
int main(){
    scanf("%d%d",&X,&k);
    for(int i=1;i<=k;i++){
        scanf("%d",&qry[i].t);
    }
    scanf("%d",&q);
    for(int i=1;i<=q;i++){
        scanf("%d%d",&qry[i+k].t,&qry[i+k].a);
        qry[i+k].id=i;
    }
    sort(qry+1,qry+q+k+1);
    down=true;
    a=0,b=X,c=0;
    for(int i=1;i<=q+k;i++){
        int x=qry[i].t-qry[i-1].t;
        if(down)update(-x);
        else update(x);
        //cout<<qry[i].t<<' '<<a<<' '<<b<<' '<<c<<endl;
        if(qry[i].id==0)down^=1;
        else ans[qry[i].id]=query(qry[i].a);
    }
    for(int i=1;i<=q;i++)printf("%d\n",ans[i]);
    return 0;
}
