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
#include <bits/stdc++.h>
#define ll long long
#define lf double
#define E complex<lf>
#define inf 0x3f3f3f3f
#define eps 1e-8
#define pa pair<int,int>
#define pb push_back
#define ms(x,y) memset(x,y,sizeof(x))
#define l(x) (x<<1)
#define r(x) (x<<1|1)
#define mod 1000000007
#define N 100010
using namespace std;
inline ll read() {
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
int n,m;
ll x,t[N],a[N],r[N];
int main() {
	x=read(),n=read();
	for (int i=1; i<=n; i++) r[i]=read();
	m=read();
	for (int i=1; i<=m; i++) t[i]=read(),a[i]=read();
	ll h=0,k=1,opt=-1,mx=x,mn=0;
	for (int i=1; i<=m; i++) {
		while (k<=n&&t[i]>=r[k]) {
			ll v=(r[k]-r[k-1])*opt;
			mx+=v,mn+=v,h+=v;
			mx=min(mx,x),mx=max(mx,0ll);
			mn=min(mn,x),mn=max(mn,0ll);
			k++,opt*=-1;
		}
		ll now=a[i]+h,d=(t[i]-r[k-1])*opt;
		now=max(now,mn),now=min(now,mx);
		now+=d,now=max(now,0ll),now=min(now,x);
		printf("%lld\n",now);
	}
}