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>
#define REP(i, a, n) for(ll i = ((ll) a); i < ((ll) n); i++)
#define INF 10000000000LL
using namespace std;
typedef long long ll;
 
ll X, K, R[200000], Q, T[200000], A[200000];
 
int main(void) {
  cin >> X >> K;
  R[0] = 0;
  REP(i, 1, K + 1) cin >> R[i];
  R[K + 1] = INF;
  cin >> Q;
  REP(i, 0, Q) cin >> T[i] >> A[i];
 
  ll lower = 0;
  ll upper = X;
 
  ll sum = 0;
  ll k = 0;
  REP(i, 0, K + 1) {
    if(i > 0) {
      lower = X - max(lower - (R[i] - R[i - 1]), 0LL);
      upper = X - max(upper - (R[i] - R[i - 1]), 0LL);
      swap(lower, upper);
    }
    sum += (i % 2 == 0 ? 1 : -1) * (R[i] - R[i - 1]);
 
    while(k < Q && R[i] <= T[k] && T[k] <= R[i + 1]) {
      ll a = i % 2 == 0 ? A[k] + sum : X - A[k] - sum;
      if(a <= lower) a = lower;
      if(a >= upper) a = upper;
      cout << (i % 2 == 0 ? max(a - (T[k] - R[i]), 0LL) : X - max(a - (T[k] - R[i]), 0LL)) << endl;
      k++;
    }
  }
}
