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 MAXN = 1e5 + 7;
int r[MAXN], t[MAXN], a[MAXN];
int main()
{
    int X, k; scanf("%d%d", &X, &k);
    for (int i = 1; i <= k; i++) scanf("%d", &r[i]);
    r[k + 1] = 1e9 + 1;
    int q; scanf("%d", &q);
    for (int i = 1; i <= q; i++) scanf("%d%d", &t[i], &a[i]);
    int i = 0, j = 1;
    int low = 0, up = X, sig = -1, bound = 0;
    int det, res;
    while (j <= q)
    {
        if (i < k && t[j] >= r[i + 1])
        {
            i++;
            det = sig * (r[i] - r[i - 1]);
            low = max(min(low + det, X), 0);
            up = max(min(up + det, X), 0);
            bound += det;
            sig *= -1; 
        }
        else
        {
            det = t[j] - r[i];
            res = max(min(a[j] + bound, up), low);
            res = max(min(X, res + sig * det), 0);
            printf("%d\n", res);
            j++;
        }
    }
    return 0;
}