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())
R = list(map(int, input().split()))
Q = int(input())


ami = [0]*(K+1)
ama = [X] + [0]*K
cusummi = [0]*(K+1)
cusumma = [0]*(K+1)
cusum = [0]*(K+1)
pr = 0
for i, r in enumerate(R):
    d = pr-r if i%2==0 else r-pr
    cusum[i+1] = cusum[i] + d
    cusummi[i+1] = min(cusummi[i], cusum[i+1])
    cusumma[i+1] = max(cusumma[i], cusum[i+1])
    ami[i+1] = min(max(ami[i] + d, 0), X)
    ama[i+1] = min(max(ama[i] + d, 0), X)
    pr = r

import bisect
for _ in range(Q):
    t, a = map(int, input().split())
    i = bisect.bisect_right(R, t)
    t1 = R[i-1] if i!=0 else 0
    t2 = t-t1
    d = -t2 if i%2==0 else t2
    if ama[i]==ami[i]:
        print(min(max(ama[i] + d, 0), X))
        continue
    ans = a+cusum[i]
    if -cusummi[i]>a:
        ans += -cusummi[i]-a
    if cusumma[i]>(X-a):
        ans -= cusumma[i]-(X-a)
    print(min(max(ans+d, 0), X))
