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 <iostream>
#include <algorithm>
using namespace std;
long long u[100010],d[100010],r[100010],mn[100010],mx[100010];
int main(){
	long long i,j,x,k,q,t,a,l;
	cin >> x >> k;
	for(i=1;i<=k;i++){
		cin >> r[i];
	}
	r[0] = 0;
	//前処理
	u[0] = 0; d[0] = x; mn[0] = 0; mx[0] = x;
	for(i=1;i<=k;i++){
		long long y = r[i]-r[i-1],k;
		if(i%2==1){
			mx[i] = max(mx[i-1]-y,(long long)0);
			mn[i] = max(mn[i-1]-y,(long long)0);
			u[i] = u[i-1]+mn[i]-(mn[i-1]-y);
			d[i] = d[i-1];
		}else{
			mx[i] = min(mx[i-1]+y,x);
			mn[i] = min(mn[i-1]+y,x);
			u[i] = u[i-1];
			d[i] = d[i-1]-(mx[i-1]+y-mx[i]);
		}
	}
		
	cin >> q;
	l = 0;
	for(j=0;j<q;j++){
		cin >> t >> a;
		for(i=l;i<=k;i++){
			if(r[i]>t){
				l = i-1;
				break;
			}
			if(i==k){
				l = i;
			}
		}
		if(u[l]>=d[l]){
			if(l%2==0){
				cout << max(mn[l]-t+r[l],(long long)0) << endl;
			}else{
				cout << min(mn[l]+t-r[l],x) << endl;
			}
		}else{
			if(a<=u[l]){
				if(l%2==0){
					cout << max(mn[l] - t + r[l],(long long)0) << endl;
				}else{
					cout << min(mn[l] + t - r[l],x) << endl;
				}
			}else if(a>=d[l]){
				if(l%2==0){
					cout << max(mx[l] - t + r[l],(long long)0) << endl;
				}else{
					cout << min(mx[l] + t - r[l],x) << endl;
				}
			}else{
				if(l%2==0){
					cout << max(mx[l]-(d[l]-a)-t+r[l],(long long)0) << endl;
				}else{
					cout << min(mx[l]-(d[l]-a)+t-r[l],x) << endl;
				}
			}
		}
	}
}