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\ninput = sys.stdin.readline\n\n"""\nX定数だった。両極端な場合が合流する。\n[0,L]上定数A、[R,X]上定数B、その間は線形\n"""\n\nX = int(input())\nK = int(input())\nR = [int(x) for x in input().split()]\nQ = int(input())\nTA = [tuple(int(x) for x in input().split()) for _ in range(Q)]\n\ntask = sorted([(r,-1) for r in R] + [(t,a) for t,a in TA])\n\nL = 0\nR = X\nA = 0\nB = X\ndx = -1 # 初めは減る方\ncurrent = 0\nanswer = []\nfor t,a in task:\n    # とりあえず上限を突破して落とす\n    A += dx * (t-current)\n    B += dx * (t-current)\n    current = t\n    if a != -1:\n        # 体積の計算\n        if a <= L:\n            x = A\n        elif a >= R:\n            x = B\n        else:\n            x = A+(B-A)//(R-L)*(a-L)\n        if x < 0:\n            x = 0\n        if x > X:\n            x = X\n        answer.append(x)\n    else:\n        dx = -dx\n        if A < B:\n            if A < 0:\n                L += (-A)\n                A = 0\n            if B > X:\n                R -= (B-X)\n                B = X\n            if A > X:\n                A = X\n                B = X\n                L = 0\n                R = 0\n            if B < 0:\n                A = 0\n                B = 0\n                L = 0\n                R = 0\n        elif A >= B:\n            if A > X:\n                L += (A-X)\n                A = X\n            if B < 0:\n                R -= (-B)\n                B = 0\n            if A < 0:\n                A = 0\n                B = 0\n                L = 0\n                R = 0\n            if B > X:\n                A = X\n                B = X\n                L = 0\n                R = 0\n        if R < L:\n            R = L\n\nprint(*answer,sep=\'\\n\')\n\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13673