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 pb push_back
#define int long long
#define mp make_pair
#define inf 998244353
#define LINF 1000000000000000007LL
#define ll long long
using namespace std;
signed main(){
	int x,k;
	vector<int> r;
	cin>>x>>k;
	vector< pair<int,int> > ev;
	for(int i=0;i<k;i++){
		int tmp;
		cin>>tmp;
		r.pb(tmp);
		ev.pb( mp ( tmp, -1 ) );
	}
	int q; cin>>q;
	for(int i=0;i<q;i++){
		int tim, a;
		cin>>tim>>a;
		ev.pb( mp( tim, a ) );
	}
	sort(ev.begin(),ev.end());
	vector<int> ans;
	int dir = -1, con = 0, ct = 0, btm = 0, tp = x;
	for(int i=0;i<ev.size();i++){
		int et = ev[i].first - ct;
		con += dir * et;
		btm = min( max( btm, -con ), x-con);
		tp = min( max( tp, -con ), x-con);
		if( ev[i].second == -1){
			dir *= -1;
		}
		else{
			int a = ev[i].second;
			a = min( max( btm, a ), tp );
			ans.pb( a + con );
		}
		ct = ev[i].first;
	}
	for(int i=0;i<ans.size();i++){
		cout<<ans[i]<<endl;
	}
}