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>

const int N = 100005;

int all, k, q, r[N], t[N], a[N];
int up, down, c;
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> all >> k;
	for (int i = 1; i <= k; ++i)
		std::cin >> r[i];
	r[0] = 0, r[k + 1] = 1e9 + 1;
	std::cin >> q; int p = 0;
	for (int i = 0; i < q; ++i)
		std::cin >> t[i] >> a[i];
	up = all; int typ = 0;
	for (int i = 0; i <= k && p < q; ++i) {
		while (p < q && t[p] >= r[i] && t[p] < r[i + 1]) {
			int x = std::min(std::max(a[p] + c, down), up), d = t[p++] - r[i];
			if (typ) x = std::min(x + d, all); else x = std::max(x - d, 0);
			std::cout << x << '\n';
		}
		int d = r[i + 1] - r[i];
		if (typ) down = std::min(down + d, all), up = std::min(up + d, all), c += d;
		else down = std::max(down - d, 0), up = std::max(up - d, 0), c -= d;
		typ ^= 1;
	}
	return 0;
}