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 REP(i,x,n) for(int i=x;i<n;i++)
#define ALL(v) (v).begin(),(v).end()
#define MP(a,b) make_pair(a,b)
typedef long long LL;
typedef pair<int, int> PI;
typedef vector<int> VI;
const LL MOD = 1000000007LL;
int r[100000];
int main() {
	int X;
	int K;
	cin >> X;
	cin >> K;
	rep(i, K) cin >> r[i];
	int Q;
	cin >> Q;
	int L = 0, R = X, C = 0;
	int j = 0;
	int prev = 0;
	rep(i, Q) {
		int t, a;
		cin >> t >> a;
		while (j < K&&r[j] <= t) {
			int T = r[j] - prev;
			if (j % 2 == 0) {
				if (T > C) L = min(L + (T - C), R);
				C = max(C - T, 0);
			}
			else {
				if (T > X - ((R - L) + C)) R = max(R - (T - (X - ((R - L) + C))), L);
				C = min(C + T, X);
			}
			prev = r[j];
			j++;
		}
		int T = t - prev;
		if (j % 2 == 0) {
			if (T > C) L = min(L + (T - C), R);
			C = max(C - T, 0);
		}
		else {
			if (T > X - ((R - L) + C)) R = max(R - (T - (X - ((R - L) + C))), L);
			C = min(C + T, X);
		}
		prev = t;
		if (a <= L) {
			cout << C << endl;
		}
		else if (R <= a) {
			cout << (R - L) + C << endl;
		}
		else {
			cout << (a - L) + C << endl;
		}
	}
}