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>

using namespace std;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int X;
	int K;
	cin >> X >> K;
	vector<int> r(K + 1);
	r[0] = 0;
	for (int i = 1; i <= K; i++) cin >> r[i];
	int Q;
	cin >> Q;
	int L = 0, R = X, c = 0;
	for (int i = 0, j = 1; i < Q; i++) {
		int t, a;
		cin >> t >> a;
		while (j <= K && t >= r[j]) {
			if (j % 2 == 1) {
				c -= r[j] - r[j - 1];
			} else {
				c += r[j] - r[j - 1];
			}
			R = max(min(R, X - c), -c);
			L = min(max(L, -c), X - c);
			j++;
		}
		int ans;
		if (a < L) {
			ans = L + c;
		} else if (a > R) {
			ans = R + c;
		} else {
			ans = a + c;
		}
		if (j % 2 == 1) ans = max(ans - t + r[j - 1], 0);
		else ans = min(ans + t - r[j - 1], X);
		cout << ans << endl;
	}
	return 0;
}