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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/**
 *    author:  otera    
**/
#include<iostream>
#include<string> 
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm> 
#include<functional>
#include<iomanip>
#include<queue>
#include<deque>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

#define int long long
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef long double ld;
const int inf=1e9+7;
const ll INF=1LL<<60 ;
const ll mod=1e9+7 ;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<int, int> P;
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

void solve() {
	int x, k; cin >> x >> k;
    vector<int> r(k + 1, 0);
    rep(i, k) {
        cin >> r[i + 1];
    }
    int q; cin >> q;
    int maa = x, mia = 0;
    int vma = x, vmi = 0;
    int cur = 0;
    rep(i, q) {
        int t, a; cin >> t >> a;
        while(cur + 1 < k and r[cur + 1] < t) {
            int ti = r[cur + 1] - r[cur];
            if(cur % 2 == 0) {
                //減
                if(vma - ti >= 0) {
                    vma -= ti;
                } else {
                    vma = 0;
                    vmi = 0;
                    mia = maa;
                    continue;
                }

                if(vmi - ti >= 0) {
                    vmi -= ti;
                } else {
                    vmi = 0;
                    mia = mia + (ti - vmi);
                }
            } else {
                //増
                if(vmi + ti <= x) {
                    vmi += ti;
                } else {
                    vmi = x;
                    vma = x;
                    mia = maa;
                    continue;
                }

                if(vma + ti <= x) {
                    vma += ti;
                } else {
                    vma = x;
                    maa = maa - (vma + ti - x);
                }
            }
            ++ cur;
        }
        if(a >= maa) {
            if(cur % 2 == 0) {
                cout << max((vma - (t - r[cur])), 0LL) << endl;
            } else {
                cout << min((vma + (t - r[cur])), x) << endl;
            }
        } else if(a <= mia) {
            if(cur % 2 == 0) {
                cout << max((vmi - (t - r[cur])), 0LL) << endl;
            } else {
                cout << min((vmi + (t - r[cur])), x) << endl;
            }
        } else {
            if(cur % 2 == 0) {
                cout << max((vmi + (a - mia) - (t - r[cur])), 0LL) << endl;
            } else {
                cout << min((vmi + (a - mia) + (t - r[cur])), x) << endl;
            }
        }
    } 
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//int t; cin >> t; rep(i, t)solve();
	solve();
    return 0;
}