#include <cstdio>

inline int Min(const int a,const int b){return a<b?a:b;}
inline int Max(const int a,const int b){return a>b?a:b;}

int X,k,q,r[100005],t[100005],a[100005],Ans;

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",&t[i],&a[i]);
    int Minn=0,Maxn=X,t1=0,t2=1,f=-1,Now=0,p=0;
    while(t2<=q)
        if(t[t2]>r[t1+1]&&t1<k)
        {
            ++t1,Now=f*(r[t1]-r[t1-1]);
            Minn=Max(0,Min(X,Minn+Now));
            Maxn=Max(0,Min(X,Maxn+Now));
            p+=Now,f*=-1;
        }
        else
        {
            Ans=Max(Minn,Min(Maxn,a[t2]+p));
            Ans=Max(0,Min(X,Ans+f*(t[t2]-r[t1])));
            printf("%d\n",Ans),++t2;
        }
    return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <cstdio>

inline int Min(const int a,const int b){return a<b?a:b;}
inline int Max(const int a,const int b){return a>b?a:b;}

int X,k,q,r[100005],t[100005],a[100005],Ans;

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",&t[i],&a[i]);
    int Minn=0,Maxn=X,t1=0,t2=1,f=-1,Now=0,p=0;
    while(t2<=q)
        if(t[t2]>r[t1+1]&&t1<k)
        {
            ++t1,Now=f*(r[t1]-r[t1-1]);
            Minn=Max(0,Min(X,Minn+Now));
            Maxn=Max(0,Min(X,Maxn+Now));
            p+=Now,f*=-1;
        }
        else
        {
            Ans=Max(Minn,Min(Maxn,a[t2]+p));
            Ans=Max(0,Min(X,Ans+f*(t[t2]-r[t1])));
            printf("%d\n",Ans),++t2;
        }
    return 0;
}

```