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 <iostream>
#include <unordered_map>
#include <algorithm>
 
#define ll long long
 
using namespace std;
int x, k, q, t, a, low, up, add, delta, now, ans, r[100007];
bool flag;
 
int cal(int tt, int lower = 0, int upper = x) {
    if (tt < lower) return lower;
    if (tt > upper) return upper;
    return tt;
}
 
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
 
    cin >> x >> k;
    for (int i = 1; i <= k; i++) cin >> r[i];
    cin >> q;
 
    low = x, now = 1;
    while (q--) {
        cin >> t >> a;
        while (t >= r[now] && now <= k) {
            delta = (flag ? 1 : -1) * (r[now] - r[now - 1]), add += delta;
            up = cal(delta + up), low = cal(delta + low);
            flag = !flag, now++;
        }
        ans = cal((flag ? 1 : -1) * (t - r[now - 1]) + cal(a + add, up, low));
        cout << ans << '\n';
    }
    return 0;
}