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 <algorithm>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>

#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> PI;
const ll mod = 1e9 + 7;



int main(void) {
  ll x;
  int k;
  cin >> x >> k;
  VL r(k + 1, 0);
  REP(i, 0, k) {
    cin >> r[i + 1];
  }
  VL el(k + 1), em(k + 1), en(k + 1);
  VL curdir(k + 1);
  el[0] = 0;
  em[0] = x;
  en[0] = 0; // max(0, min(x, a + 0))
  ll dir = -1;
  curdir[0] = -1;
  REP(i, 0, k) {
    ll pass = r[i + 1] - r[i];
    en[i + 1] = en[i] + dir * pass;
    el[i + 1] = max(0LL, min(x, el[i] + dir * pass));
    em[i + 1] = max(0LL, min(x, em[i] + dir * pass));
    dir *= -1;
    curdir[i + 1] = dir;
  }
  int q;
  cin >> q;
  REP(puella, 0, q) {
    ll t, a;
    cin >> t >> a;
    int idx = upper_bound(r.begin(), r.end(), t) - r.begin() - 1;
    ll diff = t - r[idx];
    ll base = max(el[idx], min(em[idx], a + en[idx]));
    ll ans = max(0LL, min(x, base + curdir[idx] * diff));
    cout << ans << endl;
  }
}
