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 = 'import sys\nfrom heapq import *\n\nsys.setrecursionlimit(10 ** 6)\nint1 = lambda x: int(x) - 1\np2D = lambda x: print(*x, sep="\\n")\ndef MI(): return map(int, sys.stdin.readline().split())\ndef LI(): return list(map(int, sys.stdin.readline().split()))\ndef LLI(rows_number): return [LI() for _ in range(rows_number)]\n\ndef main():\n    x = int(input())\n    k = int(input())\n    rr = LI()\n    q = int(input())\n    ta = LLI(q)\n    # [時間,命令,(a)]でタイムラインに登録\n    # 命令0でひっくり返す 1で質問に答える\n    timeline = []\n    for r in rr:\n        heappush(timeline, [r, 0, 0])\n    for t, a in ta:\n        heappush(timeline, [t, 1, a])\n    # print(timeline)\n    # Aのパーツが一度でも空や満タンになると、aは関係なくなるのが大事な所\n    # だから空になるaはal未満のaとしてひとまとめ\n    # 満タンになるaはarより大きいaとして\n    # ひとまとめで管理する\n    #       l       r\n    # a=0 1 2 3 4 5 6 7 8 9 10...\n    #       al      ar\n    # al(=2)未満のaの砂の量sl\n    # ar(=6)より大きいaの砂の量sr\n    # al以上ar以下のaの砂の量a+sm\n    sl = sm = 0\n    sr = x\n    al = 0\n    ar = x\n    rev_t = 0\n    trend = False\n    while timeline:\n        t, op, a = heappop(timeline)\n        sd = t - rev_t\n        if op:\n            if trend:\n                if a < al:\n                    print(min(sl + sd, x))\n                elif a > ar:\n                    print(min(sr + sd, x))\n                else:\n                    print(min(a + sm + sd, x))\n            else:\n                if a < al:\n                    print(max(sl - sd, 0))\n                elif a > ar:\n                    print(max(sr - sd, 0))\n                else:\n                    print(max(a + sm - sd, 0))\n        else:\n            if trend:\n                sl = min(sl + sd, x)\n                sr = min(sr + sd, x)\n                ar = min(ar, x - sm - sd - 1)\n                sm += sd\n            else:\n                sl = max(sl - sd, 0)\n                sr = max(sr - sd, 0)\n                al = max(al, sd - sm + 1)\n                sm -= sd\n            trend = not trend\n            rev_t = t\n        # print(t, op, a, al, ar, sl, sm, sr)\nmain()\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 15225