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 <vector>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int x;
    cin >> x;

    int k;
    cin >> k;

    vector<int> r(k);
    for (int i = 0; i < k; i++) {
        cin >> r[i];
    }

    int q;
    cin >> q;

    int a = 0, b = x, c = 0, e = 0;
    int i = 0, s = 0;
    for (int j = 0; j < q; j++) {
        int t, y;
        cin >> t >> y;

        for (; i < k && r[i] < t; i++) {
            int d = r[i] - s;
            s = r[i];
            if (i % 2 == 0) {
                e = max(e - d, 0);
                c -= d;
                a = max(a, 0 - c);
            } else {
                e = min(e + d, x);
                c += d;
                b = min(b, x - c);
            }
        }

        y = a < b ? (y <= a ? a : y < b ? y : b) + c : e;

        y += (t - s) * (i % 2 == 0 ? -1 : 1);
        y = min(max(y, 0), x);

        cout << y << '\n';
    }

    return 0;
}