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 <vector>
#include <algorithm>
#define rad 334
using namespace std;
int R[100005];
int K;
int X,T,A,TI;
int Q;
int sens=-1;
int ind=0;
pair<int,int> a,b;
int main()
{
    cin>>X>>K;
    a.first=b.first=0;
    a.second=b.second=X;
    for(int i=1;i<=K;i++)cin>>R[i];
    cin>>Q;
    for(int i=1;i<=Q;i++)
    {
        cin>>T>>A;
        while(ind<K&&R[ind+1]<T)
        {
            a.first+=sens*(R[ind+1]-TI);
            a.second+=sens*(R[ind+1]-TI);
            b.first+=max(0,-a.first);
            b.second-=max(0,a.second-X);
            a.first=max(0,a.first);a.first=min(X,a.first);
            a.second=max(0,a.second);a.second=min(X,a.second);
            sens*=-1;
            TI=R[ind+1];
            ind++;
        }
        a.first+=sens*(T-TI);
        a.second+=sens*(T-TI);
        b.first+=max(0,-a.first);
        b.second-=max(0,a.second-X);
        b.first=min(b.first,b.second);
        a.first=max(0,a.first);a.first=min(X,a.first);
        a.second=max(0,a.second);a.second=min(X,a.second);
        TI=T;
        cout<<min(max(A,b.first),b.second)-b.first+a.first;
        cout<<"\n";
    }
    return 0;
}
