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>
using namespace std;

typedef pair<int,int> pii;
const int N = 1e5;

int x;
int k;
int r[N+5];
int q;

vector<pii> lq;

int main(){
	scanf("%d",&x);
	scanf("%d",&k);
	for(int i=1;i<=k;i++){
		scanf("%d",&r[i]);
	}
	r[k+1] = (int) 1e9;
	scanf("%d",&q);
	for(int i=0;i<q;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		lq.push_back(make_pair(a,b));
	}
	int id = 0;
	int baw = 0;
	int at = x;
	int ps = 0;
	int maksi = x;
	int mini = 0;
	int ibaw = 0;
	int iat = x;
	for(int i=1;i<=k+1;i++){
		int dif = r[i] - r[i-1];
		while(id < lq.size() && lq[id].first <= r[i]){
			int y = lq[id].second;
			int ans = y - ps;
			if(y <= baw){
				ans = ibaw;
			}
			if(y >= at){
				ans = iat;
			}
			if(i&1){
				ans -= lq[id].first - r[i-1];
			} else {
				ans += lq[id].first - r[i-1];
			}
			ans = max(0,ans);
			ans = min(ans,x);
//			if(!(i&1)) ans = x-ans;
			printf("%d\n",ans);
			id++;
		}
		if(i&1){
			ps+=dif;
			maksi-=dif;
			mini-=dif;
			ibaw-=dif;
			iat-=dif;
			if(ibaw < 0) ibaw = 0;
			if(iat < 0) iat = 0;
			if(mini < 0) baw -= mini, mini=0;
		} else {
			ps-=dif;
			maksi+=dif;
			mini+=dif;
			ibaw+=dif;
			iat+=dif;
			if(ibaw > x) ibaw = x;
			if(iat > x) iat = x;
			if(maksi > x) at -= maksi - x,maksi = x;
		}
//		cout << "baw " << baw << " and " << at << endl;
	}
	return 0;
}

/*

0 5 9 13 14 20

1-10, 1-5,

5,6,7,8,9,10

0,1,2,3,4,5 => 4,5,6,7,8,9

*/