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;
#define rep(i,n) for(int i=0;i<n;i++)
#define int long long
int r[100006];
int t[100005], a[100005];
signed main() {
	int x, k; cin >> x >> k;
	rep(i, k)cin >> r[i+1];
	int q; cin >> q;
	rep(i, q)cin >> t[i] >> a[i];
	t[q] = 1145141919;
	r[k + 1] = 1145141919;
	int u = 0, d = x, ux = 0, dx = x;
	int now = 0;
	rep(i, k+1) {
		while (t[now] < r[i+1]) {
			int ft = a[now];
			if (ft < u)ft = u;
			else if (d < ft)ft = d;
			int D = t[now] - r[i];
			if (i & 1) {
				ft += D + ux - u;
				ft = min(ft, x);
				cout << ft << endl;
			}
			else {
				ft -= D; ft += ux - u;
				ft = max(0ll, ft);
				cout << ft << endl;
			}
			now++;
		}
		int D = r[i + 1] - r[i];
		if (i & 1) {
			ux += D; dx += D;
			if (dx > x) {
				d -= dx - x;
				dx = x;
			}
			if (ux > x) {
				u = d = x;
				ux = dx = x;
			}
		}
		else {
			ux -= D; dx -= D;
			if (ux < 0) {
				u += -ux;
				ux = 0;
			}
			if (dx < 0) {
				u = d = ux = dx = 0;
			}
		}
	}
}