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
#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	int X, K;
	cin >> X >> K;
	int r[100002];
	rep1(i, K) cin >> r[i];
	r[0] = 0;
	r[K + 1] = 2e9;

	int L = 0;
	int R = X;
	int takasa = 0;
	int katamuki = -1;
	int *ri = r;

	int Q;
	cin >> Q;
	rep(i, Q) {
		int t, a;
		cin >> t >> a;
		
		while (t > *(ri + 1)) {
			takasa += katamuki * (*(ri + 1) - *ri);
			if (katamuki < 0) {
				L -= min(0, takasa + L);
				R -= min(0, takasa + R);
			}
			else {
				L -= max(0, takasa + L - X);
				R -= max(0, takasa + R - X);
			}
			ri++;
			katamuki *= -1;
		}
		chmax(a, L);
		chmin(a, R);
		a += takasa + katamuki * (t - *ri);
		chmax(a, 0);
		chmin(a, X);
		co(a);
	}

	Would you please return 0;
}