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;

const int INF = 1000000001;
int r[100002];

static inline int magic(int l, int x, int r) {
  return max(min(r, x), l);
}

static inline void slide(int h, int dist, int &mina, int &minb, int &maxa, int &maxb) {
  minb = magic(0, minb + dist, h);
  maxb = magic(0, maxb + dist, h);
  if(maxb - minb < maxa - mina) {
    int delta = (maxa - mina) - (maxb - minb);
    if(dist < 0)
      mina += delta;
    else
      maxa -= delta;
  }
}

int main() {
  int h, n, q, t, a, mina, maxa, minb, maxb, dist, j;
#ifdef DEBUG
  freopen("codeforces.in", "r", stdin);
  freopen("codeforces.out", "w", stdout);
#endif
  scanf("%d%d", &h, &n);
  for(int i = 1; i <= n; ++i)
    scanf("%d", &r[i]);
  r[n + 1] = INF;
  n = n + 2;
  j = 1;
  mina = minb = 0;
  maxa = maxb = h;
  scanf("%d", &q);
  for(int i = 0; i < q; ++i) {
    scanf("%d%d", &t, &a);
    while(r[j] < t) {
      dist = r[j] - r[j - 1];
      if(j % 2 == 1)
        dist = -dist;
      slide(h, dist, mina, minb, maxa, maxb);
      ++j;
    }
    dist = t - r[j - 1];
    if(j % 2 == 1)
      dist = -dist;
    slide(h, dist, mina, minb, maxa, maxb);
    r[j - 1] = t;
    a = magic(mina, a, maxa);
    printf("%d\n", maxb - (maxa - a));
  }
  return 0;
}
