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

-----Sample Output-----
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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '\n"""\nWriter: SPD_9X2\nhttps://atcoder.jp/contests/arc082/tasks/arc082_d\n\naiが固定ならば、ただクエリをソートしてシミュレーションしながら答えを出せばよい\n今回はそうでないので、方法を考える\n\n一度でも砂が落ち切ってしまえば、その後はaiに関わらず、答えは等しくなる\nおちきっていない場合、常に砂が移動し続けるので、初期値からの差分が常に等しくなる\nよって、落ち切らないaの範囲を保持し、一度も落ち切らない場合の初期からの差分\n& 落ち切ってしまった場合のシミュレーション結果を用いて答えればよい\n\n→どのタイミング(区間)で落ち切るかでその後の動きが違くね？？？\n→死\n\n→aの区間は高々3つに分けられる\n→ aが少なく、a=0と同じに収束する場合、 aが大きく、a=Xと同じに収束する場合。、一度も落ち切らない場合\n→そして、a=0,a=Xと同じになる場合は、それぞれaの初期値の区間の両端から侵食していく\n→よって、a=0のシミュ、a=Xのシミュ、落ち切らない差分シミュ。その時あるaがどれに属すかの区間\nを用いてシミュレーションしながらdp的に処理してあげればよい\n\nあとは実装が面倒そう\n"""\nfrom collections import deque\nX = int(input())\nK = int(input())\n\nr = list(map(int,input().split()))\n#番兵\nr.append(float("inf"))\nr.append(0)\n\nta = deque([])\nQ = int(input())\n\nfor i in range(Q):\n    t,a = list(map(int,input().split()))\n    ta.append([t,a])\n\nZA = 0 #初期がa=0の時のシミュ結果\nXA = X #初期がXの時\nD  = 0 #差分計算\nZmax = 0 #aがZmax以下ならZAと等しくなる\nXmin = X #aがXmin以上ならXAと等しくなる\n\nfor i in range(K+1):\n\n    time = r[i] - r[i-1]\n\n    #クエリの処理(r[i]以下に関して)\n    if i % 2 == 0: #Aが減っていく\n        \n        while len(ta) > 0 and ta[0][0] <= r[i]:\n            t,a = ta.popleft()\n            td = t - r[i-1]\n\n            if a <= Zmax:\n                print((max( 0 , ZA-td )))\n            elif a >= Xmin:\n                print((max (0 , XA-td )))\n            else:\n                print((max (0 , a+D-td)))\n\n        D -= time\n        Zmax = max(Zmax,-1*D)\n        ZA = max(0,ZA-time)\n        XA = max(0,XA-time)\n\n    else: #Aが増えていく\n        \n        while len(ta) > 0 and ta[0][0] <= r[i]:\n            t,a = ta.popleft()\n            td = t - r[i-1]\n\n            if a <= Zmax:\n                print((min( X , ZA+td )))\n            elif a >= Xmin:\n                print((min (X , XA+td )))\n            else:\n                print((min (X , a+D+td)))\n\n        D += time\n        Xmin = min(Xmin,X-D)\n        ZA = min(X,ZA+time)\n        XA = min(X,XA+time)\n\n\n    \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14933