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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 100005;
typedef long long i64;

int R[MAX_N], X, K, Q;

void get(i64 &temp) {
  if (temp < 0) temp = 0;
  if (temp > X) temp = X;
}

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

  scanf("%d", &Q);

  i64 sum = 0, mul = -1;
  i64 mn = 0, mx = X;
  int l = 1;
  
  while (Q--) {
    int T;
    i64 A;
    scanf("%d%lld", &T, &A);
    while (l <= K && R[l] <= T) {
      sum += (R[l] - R[l - 1]) * mul;
      mn += (R[l] - R[l - 1]) * mul;
      mx += (R[l] - R[l - 1]) * mul;
      get(mn), get(mx), mul = -mul;
      l++;
    }
    
    A += sum;
    if (A < mn) A = mn;
    if (A > mx) A = mx;
    A += (T - R[l - 1]) * mul;
    
    get(A);
    printf("%lld\n", A);
  }
  
  return 0;
}
