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;

int X, K;
int r[100005];
int Q;

int lb[100005], ub[100005];
int sum[100005], Max[100005], Min[100005];

int main(void)
{
	cin >> X >> K;
	for(int i = 1; i <= K; i++) cin >> r[i];
	
	for(int i = 1; i <= K; i++){
		if(i % 2) sum[i] = sum[i-1] - (r[i] - r[i-1]);
		else sum[i] = sum[i-1] + (r[i] - r[i-1]);
	}
	
	Max[0] = X, ub[0] = X;
	for(int i = 1; i <= K; i++){
		ub[i] = ub[i-1];
		if(i % 2){
			Max[i] = Max[i-1] - (r[i] - r[i-1]);
			if(Max[i] < 0) Max[i] = 0;
		}
		else{
			Max[i] = Max[i-1] + (r[i] - r[i-1]);
			if(Max[i] > X){
				ub[i] -= Max[i] - X;
				Max[i] = X;
			}
		}
	}
	Min[0] = 0, lb[0] = 0;
	for(int i = 1; i <= K; i++){
		lb[i] = lb[i-1];
		if(i % 2){
			Min[i] = Min[i-1] - (r[i] - r[i-1]);
			if(Min[i] < 0){
				lb[i] += -Min[i];
				Min[i] = 0;
			}
		}
		else{
			Min[i] = Min[i-1] + (r[i] - r[i-1]);
			if(Min[i] > X) Min[i] = X;
		}
	}
	
	cin >> Q;
	int t, a, pos, val;
	for(int q = 0; q < Q; q++){
		cin >> t >> a;
		pos = upper_bound(r, r+K+1, t) - r-1;
		if(a <= lb[pos]) val = Min[pos];
		else if(a >= ub[pos]) val = Max[pos];
		else val = a + sum[pos];
		if(pos % 2){
			val += t - r[pos];
			val = min(val, X);
		}
		else{
			val -= t - r[pos];
			val = max(val, 0);
		}
		cout << val << endl;
	}
	
	return 0;
}