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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <math.h>

using namespace std;
typedef long long LL;
const int maxn = 100005;
int o[maxn],X,n,Q,t[maxn],a[maxn];
int l[maxn],fl[maxn],r[maxn];
int main()
{
	#ifdef Amberframe
		freopen("arc082f.in","r",stdin);
		freopen("arc082f.out","w",stdout);
	#endif
	scanf("%d %d",&X,&n);r[0]=X;
	for (int i=1;i<=n;i++) scanf("%d",&o[i]);
	for (int i=1;i<=n;i++)
	if (i&1) {
		int d=o[i]-o[i-1];
		if (l[i-1]==r[i-1])
			l[i]=r[i]=l[i-1],fl[i]=max(0,fl[i-1]-d);
		else if (fl[i-1]>=d)
			l[i]=l[i-1],r[i]=r[i-1],fl[i]=fl[i-1]-d;
		else if (l[i-1]+d-fl[i-1]>=r[i-1])
			l[i]=r[i]=0,fl[i]=0;
		else
			l[i]=l[i-1]+d-fl[i-1],r[i]=r[i-1],fl[i]=0;
	}
	else {
		int d=o[i]-o[i-1];
		if (l[i-1]==r[i-1])
			l[i]=r[i]=l[i-1],fl[i]=min(X,fl[i-1]+d);
		else if (fl[i-1]+r[i-1]-l[i-1]+d<=X)
			l[i]=l[i-1],r[i]=r[i-1],fl[i]=fl[i-1]+d;
		else if (fl[i-1]+d>=X)
			l[i]=r[i]=0,fl[i]=X;
		else
			l[i]=l[i-1],r[i]=r[i-1]-(fl[i-1]+r[i-1]-l[i-1]+d-X),fl[i]=fl[i-1]+d;
	}
	scanf("%d",&Q);
	for (int i=1;i<=Q;i++)
		scanf("%d %d",&t[i],&a[i]);
	for (int pos=0,i=1;i<=Q;i++)
	{
		while (pos<n&&o[pos+1]<=t[i]) ++pos;
		int cur=(a[i]<=l[pos]?fl[pos]:fl[pos]+min(r[pos],a[i])-l[pos]);
		int d=t[i]-o[pos];pos&1?cur=min(cur+d,X):cur=max(cur-d,0);
		printf("%d\n",cur);
	}
	return 0;
}