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 bisect
oX = int(input())
K = int(input())
r= list(map(int,input().split()))

# a X-a
# min(max(0,a-r[0]))
#  a-r[0]+r[1]

def calc_time(t):
    val = bisect.bisect_left(t,r)
    diff= t- r[val]
    return val,diff


# [sep1,sep2,a<sep1,sep1<a<sep2,a>sep2  ]

last =(0,oX,0 ,0,oX)
res =[last]
for i,tr in enumerate(r):
    sep1,sep2,X,Y,Z=last
    dt = tr if i==0 else tr-r[i-1]
    if i%2==0: # a ->b

        X = max(0,X-dt)
        Y = Y-dt
        Z = max(0,Z-dt)
        sep1 = max(sep1,-Y)
    else: #b-> a
        X = min(oX,X+dt)
        Y= Y+dt

        Z = min(oX,Z+dt)
        sep2 = min(sep2,oX-Y)

    last = (sep1,sep2,X,Y,Z)
    res.append(last)

#print(res)


Q = int(input())

r_index=0
r.insert(0,0)
r.append(10000000000)
for i in range(Q):
    def calc(X,r_index,diff):
        tmp = 1 if r_index%2 else -1
        return min(oX,max(0,X +tmp* diff))
    t,a =list(map(int,input().split()))
    while r[r_index]< t:
        if r[r_index+1] >t:
            break
        r_index+=1
    sep1, sep2, X, Y, Z = res[r_index]
    diff = t -r[r_index]
    #print(t,diff,r_index,r[r_index])
    if a< sep1:
        print(calc(X,r_index,diff))
    elif a>sep2:
        print(calc(Z,r_index,diff))
    else:
        print(calc(a+Y,r_index,diff))

