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>
using namespace std;
#define MAXN 100005 
#define INF 0x3f3f3f3f
#define FIO "c"
#define ll long long
#define mid ((l+r)>>1)
#define lid (id<<1)
#define rid (id<<1|1)

template <class T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	for (;!isdigit(ch) && ch!='-';ch=getchar());
	if (ch=='-')	ch=getchar(),f=-1;
	for (;isdigit(ch);ch=getchar())	x=x*10+ch-'0';
	x*=f;
}
int N,X,Q,t,a,cur=1,pre,mx,mi,r[MAXN],delta,flag=-1,ans;

inline int f(int x,int mi=0,int mx=X){return (x<mi||x>mx)?((x<mi)?mi:mx):x;}
int main()
{
//	freopen(FIO".in","r",stdin);
//	freopen(FIO".out","w",stdout);
	read(X);read(N);
	for (int i=1;i<=N;i++)	read(r[i]);
	read(Q);
	mx=X;
	while (Q--)
	{
		read(t);read(a);
		while (t>=r[cur] && cur<=N)
		{
			delta=flag*(r[cur]-r[cur-1]);pre+=delta;
			mx=f(mx+delta);mi=f(mi+delta);
			flag=-flag;cur++;
		}
	 	ans = f(flag * (t - r[cur - 1]) + f(a + pre, mi, mx));  
		printf("%d\n",ans);
	}
	fclose(stdin);fclose(stdout);
	return 0;
}

