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 pi 3.1415926535897932384626
using namespace std;
      
struct P{
  int x,y;
  bool operator < (const P &a )const{
    //if(y!=a.y)
    return y<a.y;
  };
};

int a,b,c,d,i,k,n,m,e,dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1};
int o[101101];
int l[101111][4];
int j[1];

long long x,y,z,mod=1000000007;
//string r;
char r[1];

P u[1];
queue<int> q;
//stack<P> s;

vector<int> v[1111];

bool as(P a,P b)
{
  return a.x<b.x;
}

int main()
{
  scanf("%d %d",&a,&b);
  l[0][0]=a;
  for(int t=1;t<=b+1;t++)
  {
    if(t<=b) scanf("%d",&o[t]);
    else o[t]=1000000010;
    if(t%2)
    {
      l[t][0]=max(l[t-1][0]-(o[t]-o[t-1]),0);
      l[t][1]=max(l[t-1][1]-(o[t]-o[t-1]),0);
      l[t][2]=l[t-1][2];
      if(l[t][1]==0)l[t][3]=min(a-l[t][2],l[t-1][3]-l[t-1][1]+(o[t]-o[t-1]));
      else l[t][3]=l[t-1][3];
    }
    else
    {
      l[t][0]=min(l[t-1][0]+(o[t]-o[t-1]),a);
      l[t][1]=min(l[t-1][1]+(o[t]-o[t-1]),a);
      l[t][3]=l[t-1][3];
      if(l[t][0]==a)l[t][2]=min(a-l[t][3],l[t-1][2]+(l[t-1][0]+(o[t]-o[t-1])-a));
      else l[t][2]=l[t-1][2];
    }
  }
  b++;
  scanf("%d",&c);
  for(int t=1;t<=c;t++)
  {
    scanf("%d %d",&n,&m);
    for(;o[e+1]<=n&&e<b;e++);
    int p;
    if(m<=l[e][3]) p=l[e][1];
    else if(a<=m+l[e][2]) p=l[e][0];
    else p=l[e][1]+m-l[e][3];
    if(e%2) p+=n-o[e];
    else p-=n-o[e];
    printf("%d\n",max(min(a,p),0));
  }
}
//3


