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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define MN 100000
int r[MN+5],t[MN+5],a[MN+5],s[MN+5],sl,sr,f[MN+5];
vector<int> v[MN+5];
bool cmp(int x,int y){return a[x]<a[y];}
int gf(int k){return f[k]?f[k]=gf(f[k]):k;}
int main()
{
	int x,k,q,i,j,ss=0;
	x=read();k=read();
	for(i=1;i<=k;++i)r[i]=read();
	q=read();
	for(i=1;i<=q;++i)
	{
		t[i]=read();a[i]=read();s[i]=i;
		v[upper_bound(r+1,r+k+1,t[i])-r-1].push_back(i);
	}
	sort(s+(sl=1),s+(sr=q)+1,cmp);
	for(i=0;i<v[0].size();++i)t[v[0][i]]=max(a[v[0][i]]-t[v[0][i]],0);
	for(i=1;i<=k;++i)
		if(i&1)
		{
			ss-=r[i]-r[i-1];
			if(a[s[sl]]+ss<=0)
			{
				for(j=sl;j<=sr&&a[s[j]]+ss<=0;++j);
				a[s[--j]]=-ss;
				while(sl<j)f[s[sl++]]=s[j];
			}
			for(j=0;j<v[i].size();++j)
				t[v[i][j]]=min(a[gf(v[i][j])]+ss+t[v[i][j]]-r[i],x);
		}
		else
		{
			ss+=r[i]-r[i-1];
			if(a[s[sr]]+ss>=x)
			{
				for(j=sr;j>=sl&&a[s[j]]+ss>=x;--j);
				a[s[++j]]=x-ss;
				while(sr>j)f[s[sr--]]=s[j];
			}
			for(j=0;j<v[i].size();++j)
				t[v[i][j]]=max(a[gf(v[i][j])]+ss-t[v[i][j]]+r[i],0);
		}
	for(i=1;i<=q;++i)printf("%d\n",t[i]);
}