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>

using namespace std;

int r[200000];
int X, K;

int main()
{
  int Q, T, A, minA, maxA, k = 0, t = 0, x = 0, sign = -1;

  cin >> X >> K;
  for(int i = 0; i < K; i++) {
    cin >> r[i];
  }
  r[K] = 2e9;

  minA = 0;
  maxA = X;
  cin >> Q;
  for(int i = 0; i < Q; i++) {
    cin >> T >> A;
    for(; T >= r[k]; ) {
      int diff = sign * (r[k] - t);
      x += diff;
      minA = min(X, max(0, minA + diff));
      maxA = min(X, max(0, maxA + diff));
      t = r[k];
      k++;
      sign *= -1;
    }
    T -= t;
    A = min(maxA, max(minA, A + x));
    A = min(X, max(0, A + sign * T));
    cout << A << endl;

    //cout << minA << " " << maxA << " " << x << " " << t << endl;
  }

  return 0;
}