p03616 AtCoder Regular Contest 082 - Sandglass
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
```python
X=int(input())
K=int(input())
r=list(map(int,input().split()))
Q=int(input())
p=[tuple(map(int,input().split())) for i in range(Q)]

N=Q
L=0
R=Q-1
start=0
sign="-"

# N: 処理する区間の長さ

N0 = 2**(N-1).bit_length()
data = [0]*(2*N0)
INF = 0
# 区間[l, r+1)の値をvに書き換える
# vは(t, value)という値にする (新しい値ほどtは大きくなる)
def update(l, r, v):
    L = l + N0; R = r + N0
    while L < R:
        if R & 1:
            R -= 1
            data[R-1] += v

        if L & 1:
            data[L-1] += v
            L += 1
        L >>= 1; R >>= 1
# a_iの現在の値を取得
def query(k):
    k += N0-1
    s = INF
    while k >= 0:
        s += data[k]
        k = (k - 1) // 2
    return s

q=[]
for i in range(Q):
    t,a=p[i]
    p[i]=(a,t,i)

p.sort()
dic=[-1]*Q
for i in range(Q):
    a,t,id=p[i]
    update(i,i+1,a)
    dic[id]=i
    q.append((t,id,0))

for i in range(K):
    q.append((r[i],-1,r[i]-r[i-1]*(i>0)))

q.sort()
ans=[0]*Q
for val,id,c in q:
    if id==-1:
        if sign=="-":
            update(L,R+1,-c)
            while query(L+1)<0 and L<R:
                L+=1
            a=query(L)
            if a<0:
                update(L,L+1,-a)
            start=val
            sign="+"
        else:
            update(L,R+1,c)
            while query(R-1)>X and L<R:
                R-=1
            a=query(R)
            if a>X:
                update(R,R+1,X-a)
            start=val
            sign="-"
    else:
        pid,id=id,dic[id]
        if sign=="-":
            if id<L:
                res=max(query(L)-(val-start),0)
            elif id>R:
                res=max(query(R)-(val-start),0)
            else:
                res=max(query(id)-(val-start),0)
        else:
            if id<L:
                res=min(query(L)+(val-start),X)
            elif id>R:
                res=min(query(R)+(val-start),X)
            else:
                res=min(query(id)+(val-start),X)
        ans[pid]=res


for i in range(Q):
    print(ans[i])

```