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 <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
#include <cstring>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
#define CLR(mat) memset(mat, 0, sizeof(mat))
typedef long long ll;
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll X, K; cin >> X >> K;
  ll r[K+1]; FOR(i,1,K+1) cin >> r[i];
  r[0] = 0;
  ll Q; cin >> Q;
  ll mx = X, mn = 0;
  int j = 1;
  ll d = 0;
  FOR(i,0,Q) {
    ll t, a; cin >> t >> a;
    while(j <= K && r[j] <= t) {
      if(j % 2 == 1) {
        mx = max(0LL, mx - (r[j] - r[j-1]));
        mn = max(0LL, mn - (r[j] - r[j-1]));
        d -= (r[j] - r[j-1]);
      } else {
        mx = min(X, mx + (r[j] - r[j-1]));
        mn = min(X, mn + (r[j] - r[j-1]));
        d += (r[j] - r[j-1]);
      }
      j++;
    }
    ll dd = min(mx, max(mn, a + d)) + (j % 2 == 1 ? - (t - r[j-1]) : (t - r[j-1]));
    cout << max(min(dd, X), 0LL) << endl;
  }
  return 0;
}