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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define EACH(i, c) for (auto &(i): c)
#define all(c) begin(c),end(c)
#define EXIST(s, e) ((s).find(e)!=(s).end())
#define SORT(c) sort(begin(c),end(c))
#define pb emplace_back
#define MP make_pair
#define SZ(a) int((a).size())

//#define LOCAL 0
//#ifdef LOCAL
//#define DEBUG(s) cout << (s) << endl
//#define dump(x)  cerr << #x << " = " << (x) << endl
//#define BR cout << endl;
//#else
//#define DEBUG(s) do{}while(0)
//#define dump(x) do{}while(0)
//#define BR
//#endif


//改造
typedef long long int ll;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0 //デバッグする時1にしてね
#define dump(x)  cerr << #x << " = " << (x) << endl
#define MOD 1000000007
//ここから編集する


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll X;
    cin >> X;
    int K;
    cin >> K;
    vector<ll> r(K);
    r.push_back(0);
    rep(i, 0, K) cin >> r[i + 1];

    //Cr[i]を求める
    vector<ll> Cr(K + 1);
    for (int i = 0; i < K; i++) {
        if (i % 2 == 0) {
            Cr[i + 1] = max(0LL, Cr[i] - (r[i + 1] - r[i]));
        } else {
            Cr[i + 1] = min(X, Cr[i] + (r[i + 1] - r[i]));
        }
    }

    //BEr[i],ENr[i]を求める
    vector<ll> BEr(K + 1), ENr(K + 1);
    BEr[0] = 0ll;
    ENr[0] = X;
    for (int i = 0; i < K; i++) {
        if (i % 2 == 0 && Cr[i] - (r[i + 1] - r[i]) < 0) {
            BEr[i + 1] = max((r[i + 1] - r[i]) - Cr[i], BEr[i]);
        } else {
            BEr[i + 1] = BEr[i];
        }
    }

    for (int i = 0; i < K; i++) {
        if (i % 2 == 1 && Cr[i] + (r[i + 1] - r[i]) + ENr[i] > X) {
            ll tmp = (ENr[i] - BEr[i]) + Cr[i] + (r[i + 1] - r[i]);
            ENr[i + 1] = min(2 * X - tmp, ENr[i]);
        } else {
            ENr[i + 1] = ENr[i];
        }
    }


    int Q;
    cin >> Q;
    while (Q--) {
        ll t, a = 0;
        cin >> t >> a;
        ll A0 = a;

        int tpoint = upper_bound(all(r), t) - r.begin() - 1;
        ll BEt = BEr[tpoint];
        ll ENt = ENr[tpoint];
        ll Ct;

        if (tpoint % 2 == 0) {
            if (Cr[tpoint] - (t - r[tpoint]) < 0) {
//                BEt = max(BEt, (t - (r[tpoint] - r[tpoint])));
                BEt = max(BEt, (t - r[tpoint]) - Cr[tpoint]);
            }
            Ct = max(0LL, Cr[tpoint] - (t - r[tpoint]));
        } else {
            if (Cr[tpoint] + (t - r[tpoint]) < 0) {
//                ENt = min(ENt, X - Cr[tpoint] - (t - r[tpoint]));
                ENt = min(ENt, Cr[tpoint] + (t - r[tpoint]));
            }
            Ct = min(X, Cr[tpoint] + (t - r[tpoint]));
        }
        ll At;
        if (A0 < BEt) {
            At = Ct;
        } else if (ENt < A0) {
            At = Ct + (ENt - BEt);
        } else {
            At = Ct + (A0 - BEt);
        }

        cout << At << endl;
    }


    return 0;
}