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;

struct Event {
  int t;
  int d;
  int a;
  bool operator<(const Event& r) const {
    return t < r.t;
  }
};

int R, K, Q;
int s[100000];
int t[100000];
int a[100000];

void sub(int t, int d, int a) {
  static int prev_d = 1;
  static int prev_t = 0;
  static int l = 0;
  static int r = R;
  static int x = 0;
  
  bool q = d == 0;
  if (q) {
    d = -prev_d;
  }
  
  x += d * (t - prev_t);
  l = min(max(l, -x), -x + R);
  r = min(max(r, -x), -x + R);
  
  if (q) {
    a = max(a, l);
    a = min(a, r);
    cout << x + a << endl;
  }
  
  if (!q) {
    prev_d = d;
  }
  prev_t = t;
}

int main() {
  cin >> R >> K;
  for (int i = 0; i < K; ++i) cin >> s[i];
  cin >> Q;
  for (int i = 0; i < Q; ++i) cin >> t[i] >> a[i];
  
  vector<Event> es;
  int d = -1;
  for (int i = 0; i < K; ++i) {
    es.push_back(Event{ s[i], d, 0 });
    d = -d;
  }
  for (int i = 0; i < Q; ++i) {
    es.push_back(Event{ t[i], 0, a[i] });
  }
  sort(es.begin(), es.end());
  for (auto e : es) {
    sub(e.t, e.d, e.a);
  }
}