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 Ruby (2.3.3) (judge verdict: Wrong Answer)

x = gets.to_i
k = gets.to_i
r = [0] + gets.split.map(&:to_i)
q = gets.to_i
t, a = [], []
q.times do |i|
    t[i], a[i] = gets.split.map(&:to_i)
end

a_is_upside = true

1.upto(k) do |k_i|
    break if a.empty?
    if t.first < r[k_i]
        if a_is_upside
            p [a.first - (t.first - r[k_i - 1]), 0].max
        else
            p [a.first + (t.first - r[k_i - 1]), x].min
        end
        a.shift
        t.shift
        redo
    end
    dr = r[k_i] - r[k_i - 1]
    a.each_with_index do |aa, ai|
        if a_is_upside
            if aa <= dr
                a[ai] = 0
            else
                a[ai] -= dr
            end
        else
            if aa >= x - dr
                a[ai] = x
            else
                a[ai] += dr
            end
        end
    end
    a_is_upside = !a_is_upside
end

a.each_with_index do |aa, ai|
    if a_is_upside
        p [a.first - (t.first - r.last), 0].max
    else
        p [a.first + (t.first - r.last), x].min
    end
end