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;


#define MAXN 100002
typedef long long ll;
ll sum[MAXN];
int r[MAXN], lo[MAXN], hi[MAXN];
int x, k;
void solve(int t, int a) {
  int i = upper_bound(r+1, r+k+2, t) - r - 1;
  int ans = (i%2 ? 1 : -1)*(t-r[i]);
  if (a<=lo[i]-sum[i]) {
    ans += lo[i];
  } else if (a<=hi[i]-sum[i]) {
    ans += sum[i] + a;
  } else {
    ans += hi[i];
  }
  cout << min(max(ans, 0), x) << endl;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> x >> k;
  for (int i = 1; i <= k; ++i) {
    cin >> r[i];
    sum[i] += sum[i-1] + ((i%2) ? -1 : 1)*(r[i]-r[i-1]);
  }
  r[k+1] = 2e9;
  hi[0] = x;
  for (int i = 1; i <= k+1; ++i) {
    if (i%2) {
      lo[i] = max(0, lo[i-1]-r[i]+r[i-1]);
      hi[i] = max(0, hi[i-1]-r[i]+r[i-1]);
    } else {
      lo[i] = min(x, lo[i-1]+r[i]-r[i-1]);
      hi[i] = min(x, hi[i-1]+r[i]-r[i-1]);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int t, a;
    cin >> t >> a;
    solve(t, a);
  }
  return 0;
}
