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>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<functional>
#include<tuple>
using namespace std;
int X, K, Q, p[100009], ans[100009], L = 0, R, s = 0, t; bool unity = false;
int l[100009], r[100009], v[100009]; bool u[100009];
int main() {
	cin >> X >> K; R = X; t = X;
	for (int i = 1; i < 100009; i++)p[i] = 1000000009;
	for (int i = 1; i <= K; i++)cin >> p[i]; p[K + 1] = 1000000007; K += 2;
	l[0] = 0; r[0] = R; v[0] = 0;
	for (int i = 1; i < K; i++) {
		if (i % 2 == 1) { s -= (p[i] - p[i - 1]); t -= (p[i] - p[i - 1]); if (s < 0) { L += (-s); s = 0; }if (t < 0) { unity = true; } }
		if (i % 2 == 0) { s += (p[i] - p[i - 1]); t += (p[i] - p[i - 1]); if (s > X) { s = X; unity = true; }if (t > X) { R -= (t - X); t = X; } }
		if (L >= R)unity = true;
		u[i] = unity; l[i] = L; r[i] = R; v[i] = s;
	}
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		int p1, p2; cin >> p1 >> p2;
		int pos1 = upper_bound(p, p + K, p1) - p; pos1--;
		int GG = max(l[pos1], min(r[pos1], p2)) - l[pos1]; GG += v[pos1];
		if (pos1 % 2 == 0)GG -= (p1 - p[pos1]); else GG += (p1 - p[pos1]);
		GG = max(0, min(X, GG));
		cout << GG << endl;
	}
	return 0;
}