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 <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
using namespace std;
using LL = long long;
constexpr int MOD = 1000000000 + 7;
constexpr int INF = 2000000000;
const double PI = acos(-1);

int main() {
	int x, k;
	cin >> x >> k;
	int r[100000];
	for (int i = 0; i < k; i++) {
		cin >> r[i];
	}
	int q;
	cin >> q;
	int t[100000], a[100000];
	for (int i = 0; i < q; i++) {
		cin >> t[i] >> a[i];
	}
	int L = 0, R = x;
	int j = 0, p = 0, d = 0;
	int lv = 0, uv = x;
	for (int i = 0; i < q; i++) {
		while (j < k&&r[j] <= t[i]) {
			if (j % 2 == 0) {
				p += r[j] - d;
				lv = max(lv - (r[j] - d), 0);
				uv = max(uv - (r[j] - d), 0);
			}
			else {
				p -= r[j] - d;
				lv = min(lv + (r[j] - d), x);
				uv = min(uv + (r[j] - d), x);
			}
			L = max(L, p);
			R = min(R, x + p);
			d = r[j++];
		}
		int c;
		if (L <= a[i] && a[i] <= R) c = a[i] - p;
		if (a[i]<L) c = lv;
		if (R<a[i]) c = uv;
		if (j % 2 == 0) c = max(c - (t[i] - d), 0);
		else c = min(c + (t[i] - d), x);
		cout << c << endl;
	}
}
