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.

Examples

Input

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


Output

60
1
120


Input

100
1
100000
4
0 100
90 100
100 100
101 100


Output

100
10
0
0


Input

100
5
48 141 231 314 425
7
0 19
50 98
143 30
231 55
342 0
365 100
600 10


Output

19
52
91
10
58
42
100
import sys
input = sys.stdin.readline

x = int(input())
k = int(input())
R = [(_r, i%2) for i, _r in enumerate(map(int, input().split()))]
n = int(input())
TA = [list(map(int, input().split())) for _ in range(n)]
T, A = zip(*TA)
L = [None]*n
b = 0
for i, t in enumerate(T, 1):
  R.append((t, -i))
R.sort()
dx = 1
l, r, p, q = 0, x, 0, x
for _r, i in R:
  d = _r-b
  b = _r
  if dx == 1:
    if d <= p:
      p -= d
      q -= d
    elif d < q:
      q -= d
      l = r-q
      p = 0
    else:
      l = 0
      r = 0
      p = 0
      q = 0
  else:
    if d <= x-q:
      p += d
      q += d
    elif d < x-p:
      p += d
      r = l+(x-p)
      q = x
    else:
      l = x
      r = x
      p = x
      q = x
  if i >= 0:
    dx = i
  else:
    a = A[-i-1]
    if a < l:
      res = p
    elif a <= r:
      res = (a-l)+p
    else:
      res = q
    L[-i-1] = res
print(*L, sep="\n")