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, Q;
  int R[100001];


  scanf("%d %d", &X, &K);
  for(int i = 1; i <= K; i++) scanf("%d", &R[i]);
  scanf("%d", &Q);

  R[0] = 0;
  R[K + 1] = 1 << 30;

  int tail = 1;
  int low = 0, high = X;
  int latte = 0, malta = X;

  while(Q--) {
    int T, A;
    scanf("%d %d", &T, &A);
    while(R[tail] <= T) {
      auto beet = R[tail] - R[tail - 1];
      auto delta = beet * (tail % 2 == 0 ? 1 : -1);
      if(tail % 2 == 0) high -= max(0, malta + delta - X);
      else low += max(0, -(latte + delta));
      latte = min(X, max(0, latte + delta));
      malta = min(X, max(0, malta + delta));
      ++tail;
    }
    auto beet = T - R[tail - 1];
    auto delta = beet * (tail % 2 == 0 ? 1 : -1);
    if(low <= A && A <= high) printf("%d\n", min(X, max(0, latte + delta + (A - low))));
    else if(A < low) printf("%d\n", min(X, max(0, latte + delta)));
    else printf("%d\n", min(X, max(0, malta + delta)));
  }

}