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
//f(t, a) (t固定) : ____/----の形をしているから, (__の値b, --の値c, __の終わりd, --の始まりe)を適当に更新する.
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

int x;
int k;
int r[100001];
int q;
int t[100000], a[100000];

int h(int i) {
	if (i & 1) return -1;
	return 1;
}

signed main() {
	int i;
	cin >> x >> k;
	for (i = 1; i <= k; i++) cin >> r[i];
	
	int b = 0, c = x, d = 0, e = x, j = 1;
	cin >> q;
	for (i = 0; i < q; i++) {
		cin >> t[i] >> a[i];
		for (; j <= k && r[j] <= t[i]; j++) {
			if (h(j) == -1) d = d + max(0LL, r[j] - r[j - 1] - b);
			b = min(x, max(0LL, b + h(j) * (r[j] - r[j - 1])));
			c = min(x, max(0LL, c + h(j) * (r[j] - r[j - 1])));
			e = d + (c - b);
		}
		
		int ans;
		if (a[i] <= d) ans = b;
		else if (a[i] <= e) ans = b + (a[i] - d);
		else ans = c;
		ans = min(x, max(0LL, ans + h(j) * (t[i] - r[j - 1])));
		cout << ans << endl;
	}
	return 0;
}