Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

180
3
60 120 180
3
30 90
61 1
180 180

Sample Output 1

60
1
120

In the first query, 30 out of the initial 90 grams of sand will drop from bulb A, resulting in 60 grams.
In the second query, the initial 1 gram of sand will drop from bulb A, and nothing will happen for the next 59 seconds. Then, we will turn over the sandglass, and 1 second after this, bulb A contains 1 gram of sand at the time in question.

Sample Input 2

100
1
100000
4
0 100
90 100
100 100
101 100

Sample Output 2

100
10
0
0

In every query, the upper bulb initially contains 100 grams, and the question in time comes before we turn over the sandglass.

Sample Input 3

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10

Sample Output 3

19
52
91
10
58
42
100

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h> 
 
using namespace std;
 

 
//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
typedef long long LL;

//container util
//------------------------------------------
#define PB emplace_back
#define MP make_pair 
#define SZ(a) int((a).size())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SORT(c) sort((c).begin(),(c).end())
//constant
//--------------------------------------------
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
 
const LL INF=LONG_MAX;
const double EPS=0.00001;
const long double PI = acos(-1.0);


int main(){
    LL X,K,Q;
    cin>>X>>K;
    vector<LL>r(K);
    REP(i,K)cin>>r[i];
    cin>>Q;
    
    REP(i,Q){
        LL t,a;
        cin>>t>>a;
        auto itr=lower_bound(r.begin(),r.end(),t);
        int pos=itr-r.begin();
        LL ue=a;
        REP(j,pos){
            if(j==0)ue=max(0LL,ue-r[j]);
            else ue=max(0LL,ue-(r[j]-r[j-1]));
            ue=X-ue;
        }
        if(pos!=0)t-=r[pos-1];
        ue=max(0LL,ue-t);
        if(pos%2)cout<<X-ue<<endl;
        else cout<<ue<<endl;
    }
    return 0;
}
/*
100
5
48 141 231 314 425
2
143 30
342 0
*/