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
X = int(input())
K = int(input())
rs = list(map(int,input().split()))
Q = int(input())
qs = [tuple(map(int,input().split())) for i in range(Q)]

dx = -1
ri = 0
offset = 0
upper,lower = X,0
prev_r = 0
ans = []
for t,a in qs:
    while ri < len(rs) and rs[ri] <= t:
        tmp_offset = dx * (rs[ri] - prev_r)
        if dx == 1:
            upper = min(X, upper + tmp_offset)
            lower = min(X, lower + tmp_offset)
        else:
            upper = max(0, upper + tmp_offset)
            lower = max(0, lower + tmp_offset)
        offset += tmp_offset
        dx *= -1
        prev_r = rs[ri]
        ri += 1
    a = max(lower, min(upper, a+offset))
    dt = t - prev_r
    a = max(0, min(X, a + dx*dt))
    ans.append(a)

print(*ans, sep='\n')