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
ri = lambda: map(int, raw_input().split())

x = input()
k, r = input(), ri()

j, v, lb, rb, pt = 0, 0, 0, x, 0
L, R = 0, x
for i in xrange(input()):
    t, a = ri()
    while j < k and r[j] <= t:
        d = r[j] - pt
        pt = r[j]
        if j % 2 == 0:
            v -= d
            lb = max(lb - d, 0)
            rb = max(rb - d, 0)
        else:
            v += d
            lb = min(lb + d, x)
            rb = min(rb + d, x)
        L = max(L, -v)
        R = min(R, x - v)
        j += 1
    ans = 0
    if L <= a <= R:
        ans = a + v
    else:
        ans = lb if a < L else rb
    if j % 2 == 0:
        ans = max(ans - (t - pt), 0)
    else:
        ans = min(ans + (t - pt), x)
    print ans