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
from heapq import *

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]

def main():
    x = int(input())
    k = int(input())
    rr = LI()
    q = int(input())
    ta = LLI(q)
    # [時間,命令,(a)]でタイムラインに登録
    # 命令0でひっくり返す 1で質問に答える
    timeline = []
    for r in rr:
        heappush(timeline, [r, 0, 0])
    for t, a in ta:
        heappush(timeline, [t, 1, a])
    # print(timeline)
    # Aのパーツが一度でも空や満タンになると、aは関係なくなるのが大事な所
    # だから空になるaはal未満のaとしてひとまとめ
    # 満タンになるaはarより大きいaとして
    # ひとまとめで管理する
    #       l       r
    # a=0 1 2 3 4 5 6 7 8 9 10...
    #       al      ar
    # al(=2)未満のaの砂の量sl
    # ar(=6)より大きいaの砂の量sr
    # al以上ar以下のaの砂の量a+sm
    sl = sm = 0
    sr = x
    al = 0
    ar = x
    rev_t = 0
    trend = False
    while timeline:
        t, op, a = heappop(timeline)
        sd = t - rev_t
        if op:
            if trend:
                if a < al:
                    print(min(sl + sd, x))
                elif a > ar:
                    print(min(sr + sd, x))
                else:
                    print(min(a + sm + sd, x))
            else:
                if a < al:
                    print(max(sl - sd, 0))
                elif a > ar:
                    print(max(sr - sd, 0))
                else:
                    print(max(a + sm - sd, 0))
        else:
            if trend:
                sl = min(sl + sd, x)
                sr = min(sr + sd, x)
                ar = min(ar, x - sm - sd - 1)
                sm += sd
            else:
                sl = max(sl - sd, 0)
                sr = max(sr - sd, 0)
                al = max(al, sd - sm + 1)
                sm -= sd
            trend = not trend
            rev_t = t
        # print(t, op, a, al, ar, sl, sm, sr)
main()
