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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// a == -1 is a turn query
struct Query {
  int t, a;

  Query (int _t, int _a) : t(_t), a(_a) {}
};

bool operator< (Query p, Query q) {
  return p.t < q.t;
}

int main () {
  ios::sync_with_stdio(false);

  int cap;
  cin >> cap;

  int K;
  cin >> K;

  vector<Query> queries;
  for (int i = 0; i < K; i++) {
    int r;
    cin >> r;

    queries.push_back(Query(r, -1));
  }

  int Q;
  cin >> Q;

  for (int i = 0; i < Q; i++) {
    int t, a;
    cin >> t >> a;

    queries.push_back(Query(t, a));
  }

  sort(queries.begin(), queries.end());

  int speed = -1, last = 0, l = 0, r = cap, add = 0;
  for (Query q : queries) {
    add += speed * (q.t - last);
    last = q.t;
    
    if (speed == -1) {
      l = max(l, -add);
      if (l > r) r = l;
    } else {
      r = min(r, cap - add);
      if (r < l) l = r;
    }

    if (q.a == -1) {
      speed *= -1;
    } else {
      q.a = min(q.a, r);
      q.a = max(q.a, l);
      cout << q.a + add << '\n';
    }
  }
}
