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;
const int MAX = (1e5)+1;

int X, n, q;
long long r[MAX], t[MAX], a[MAX];
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> X >> n;
	for(int i = 1; i <= n; i++) cin >> r[i];
	cin >> q;
	long long mn = 0, mx = X, sum = 0;
	for(int i = 1, j = 1, mul = -1; i <= q; i++) {
		cin >> t[i] >> a[i];
		while(j <= n && r[j] <= t[i]) {
			sum += mul * (r[j] - r[j-1]);
			mn += mul * (r[j] - r[j-1]);
			mx += mul * (r[j] - r[j-1]);
			mn = max(0ll, mn);
			mn = min(1ll * X, mn);
			mx = max(0ll, mx);
			mx = min(1ll * X, mx);
			mul *= -1;
			++j;
		}
		
		a[i] += sum;
		if(a[i] < mn) a[i] = mn;
		if(a[i] > mx) a[i] = mx;
		a[i] += mul * (t[i] - r[j-1]); // always over 1
		a[i] = max(0ll, a[i]);
		a[i] = min(1ll * X, a[i]);

		cout << a[i] << '\n';
	}
}
