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;
const int N=1e5+10;
int zs,ctw,n,q,nt,nv,nfx=1,nzl,xyz,fzt[N],fw1[N],fw2[N];
void mchange()
{
    int kbl=fzt[ctw+1]-fzt[ctw];
    if(nfx==1)
    {
        fw1[ctw+1]=max(0,fw1[ctw]-kbl);
        fw2[ctw+1]=max(0,fw2[ctw]-kbl);
        nzl-=kbl; 
    }
    else
    {
        fw1[ctw+1]=min(zs,fw1[ctw]+kbl);
        fw2[ctw+1]=min(zs,fw2[ctw]+kbl);
        nzl+=kbl;
    }
    nfx=nfx*(-1); 
}
int main()
{
    //freopen("c4.in","r",stdin);
    //freopen("test.out","w",stdout);
    int i,j;
    scanf("%d",&zs);
    scanf("%d",&n);
    for(i=1;i<=n;++i)
    {
        scanf("%d",&fzt[i]);
    }
    scanf("%d",&q);
    fw2[0]=zs; 
    for(i=1;i<=q;++i)
    {
        scanf("%d%d",&nt,&nv);
        while(ctw<n&&fzt[ctw+1]<nt) 
        {
           mchange();
           ctw++;
        }
        if(nv+nzl>=fw1[ctw]&&nv+nzl<=fw2[ctw])	xyz=nv+nzl;
        if(nv+nzl<fw1[ctw]) xyz=fw1[ctw];
        if(nv+nzl>fw2[ctw]) xyz=fw2[ctw];
        int kbl=nt-fzt[ctw];
        if(nfx==1) xyz=max(0,xyz-kbl);
        else xyz=min(zs,xyz+kbl);
        printf("%d\n",xyz);
    }
    return 0;
}