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;

int main() {
  int X, K;
  cin >> X >> K;
  vector<int> r(K + 1); r[0] = 0;
  for (int i = 0; i < K; i++) cin >> r[i + 1];
  int Q;
  cin >> Q;
  vector<int> t(Q), a(Q);
  for (int i = 0; i < Q; i++) cin >> t[i] >> a[i];

  int now = 0, pos = 0;
  int L = 0, R = X;
  int va = 0, vb = X;
  for (int i = 0; i < Q; i++) {
    for (int j = now + 1; j <= K; j++) {
      if (r[j] > t[i]) break;
      if (j % 2 == 1) {
        pos += r[j] - r[j - 1];
        va = max(0, va - (r[j] - r[j - 1]));
        vb = max(0, vb - (r[j] - r[j - 1]));
      } else {
        pos -= r[j] - r[j - 1];
        va = min(X, va + (r[j] - r[j - 1]));
        vb = min(X, vb + (r[j] - r[j - 1]));
      }
      L = max(L, pos);
      R = min(R, X + pos);
      now++;
    }
    int c;
    if (L <= a[i] && a[i] <= R) c = a[i] - pos;
    if (a[i] < L) c = va;
    if (R < a[i]) c = vb;
    c = (now % 2 == 0) ? max(0, c - (t[i] - r[now])) : min(X, c + (t[i] - r[now]));
    cout << c << endl;
  }
}
