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 Nmax = 1e5 + 5;
int X, n, a[Nmax], mode, tmp, t, A, B, C, init, i, q;

void advance(int x)
{
    if(mode == -1)
    {
        int t = A + C;
        C -= x;
        if(t <= x) A += x-t;
    }
    else
    {
        int t = X - (B + C);
        C += x;
        if(t <= x) B -= x-t;
    }

    if(A > B)
    {
        if(mode == 1) A = B = X, C = 0;
            else A = B = 0, C = 0;
    }
}

int f(int x)
{
    if(x <= A) return A + C;
    if(x >= B) return B + C;
    return x + C;
}

int main()
{
   // freopen("input", "r", stdin);

    cin >> X >> n;
    for(i=1; i<=n; ++i) cin >> a[i];

    cin >> q; i = 1; A = 0, B = X, C = 0;
    tmp = 0; mode = -1;

    while(q--)
    {
        cin >> t >> init;
        while(i<=n && a[i] <= t)
        {
            advance(a[i] - tmp);
            mode = -mode; tmp = a[i++];
        }

        advance(t - tmp); tmp = t;
        cout << f(init) << '\n';
    }

    return 0;
}
