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
/* --- author: dxm --- */
#include<bits/stdc++.h>
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define fst first
#define snd second
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
#define DEBUG

const int maxn=100005;
int x,k,q;
int a[maxn],b[maxn],c[maxn],r[maxn];

int main(){
#ifndef DEBUG
	freopen("my.in","r",stdin);
	freopen("my.out","w",stdout);
#endif
	scanf("%d%d",&x,&k);
	REP(i,k)scanf("%d",&r[i]);
	b[0]=x;
	REP(i,k){
		a[i]=a[i-1];b[i]=b[i-1];
		if(i&1){
			a[i]=max(a[i],min(r[i]-r[i-1]-c[i-1],b[i]));
			c[i]=max(a[i-1]+c[i-1]-r[i]+r[i-1],0)-a[i];
		}
		else{
			b[i]=min(b[i],max(x-(r[i]-r[i-1])-c[i-1],a[i]));
			c[i]=min(b[i-1]+c[i-1]+r[i]-r[i-1],x)-b[i];
		}
//		cout<<a[i]<<" "<<b[i]<<" "<<c[i]<<endl;
	} 
	scanf("%d",&q);
	while(q--){
		int t,ini;
		scanf("%d%d",&t,&ini);
		int pos=upper_bound(r+1,r+1+k,t)-r-1,ans=c[pos];
		if(ini<=a[pos])ans+=a[pos];
		else if(ini>=b[pos])ans+=b[pos];
		else ans+=ini;
		if(pos&1)ans=min(x,ans+(t-r[pos]));
		else ans=max(0,ans-(t-r[pos]));
		printf("%d\n",ans);
	}
	return 0;
}

/*
Input:
-----------------
Output:
*/