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 rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

ll T;
int K,Q;

int main(){
	cin>>T;
	cin>>K;
	set<P> events;
	rep(i,K){
		int t;
		cin>>t;
		events.insert(P(t,-1));	//reverse query
	}
	cin>>Q;
	rep(i,Q){
		ll t,f;
		cin>>t>>f;
		events.insert(P(t,f));
	}

	bool inc = 0;
	ll pt = 0;
	ll a = 0, b = T, c = 0;
	for(P e:events){
		ll t = e.fs;
		ll x = e.sc;

		if(inc){
			ll d = t-pt;
			c += d;
			if(b+c>T) b = T-c;
			if(a>b) a = b;
		}else{
			ll d = t-pt;
			c -= d;
			if(a+c<0) a = -c;
			if(a>b) b = a;
		}
		if(x==-1){	//reverse
			inc = !inc;
		}else{		//answer
			cout<<max(a,min(b,x))+c<<endl;
		}
		pt = t;
	}
}
