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定数だった。両極端な場合が合流する。
[0,L]上定数A、[R,X]上定数B、その間は線形
"""

X = int(input())
K = int(input())
R = [int(x) for x in input().split()]
Q = int(input())
TA = [tuple(int(x) for x in input().split()) for _ in range(Q)]

task = sorted([(r,-1) for r in R] + [(t,a) for t,a in TA])

L = 0
R = X
A = 0
B = X
dx = -1 # 初めは減る方
current = 0
answer = []
for t,a in task:
    # とりあえず上限を突破して落とす
    A += dx * (t-current)
    B += dx * (t-current)
    current = t
    if a != -1:
        # 体積の計算
        if a <= L:
            x = A
        elif a >= R:
            x = B
        else:
            x = A+(B-A)//(R-L)*(a-L)
        if x < 0:
            x = 0
        if x > X:
            x = X
        answer.append(x)
    else:
        dx = -dx
        if A < B:
            if A < 0:
                L += (-A)
                A = 0
            if B > X:
                R -= (B-X)
                B = X
            if A > X:
                A = X
                B = X
                L = 0
                R = 0
            if B < 0:
                A = 0
                B = 0
                L = 0
                R = 0
        elif A >= B:
            if A > X:
                L += (A-X)
                A = X
            if B < 0:
                R -= (-B)
                B = 0
            if A < 0:
                A = 0
                B = 0
                L = 0
                R = 0
            if B > X:
                A = X
                B = X
                L = 0
                R = 0
        if R < L:
            R = L

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

