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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int X, K, Q, v[100009], t[100009], a[100009], ans[100009]; vector<int> g[100009];
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	cin >> X >> K; K++;
	for (int i = 1; i < K; i++) cin >> v[i];
	cin >> Q;
	for (int i = 0; i < Q; i++) {
		cin >> t[i] >> a[i];
		g[lower_bound(v, v + K, t[i] + 1) - v - 1].push_back(i);
	}
	int l = 0, r = X, e = 0;
	for (int i = 0; i < K; i++) {
		for (int j : g[i]) {
			int val = (a[j] < l ? e : (a[j] > r ? e + r - l : e + a[j] - l));
			ans[j] = (i & 1 ? min(X, val + t[j] - v[i]) : max(0, val - t[j] + v[i]));
		}
		if (i & 1) {
			e += v[i + 1] - v[i];
			if (e + r - l > X) {
				r -= e + r - l - X;
				if (r < l) r = l;
				e = X - r + l;
			}
		}
		else {
			e -= v[i + 1] - v[i];
			if (e < 0) {
				l -= e; e = 0;
				if (l > r) l = r;
			}
		}
	}
	for (int i = 0; i < Q; i++) cout << ans[i] << '\n';
	return 0;
}