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;
#define LL long long
int X;
int n, q;
const int N = 200000;
int ri[N];
struct P3
{
    LL a, b, c;
} L[N];
P3 change(P3 a, int d)
{
    if (d <= 0)
    {
        if (a.c + d >= 0) return (P3){a.a, a.b, a.c + d};
        else if (a.c + a.b + d >= 0) return (P3){a.a + (-d) - a.c, a.b + a.a - (a.a + (-d) - a.c), 0};
        else return (P3){X, 0, 0};
    }
    else
    {
        if (a.c + a.b + d <= X) return (P3){a.a, a.b, a.c + d};
        else if (a.c + d <= X) return (P3){a.a, X - d - a.c, a.c + d};
        else return (P3){X, 0, X};
    }
}
int main()
{
    cin >> X >> n;
    L[0] = (P3){0, X, 0};
    for (int i = 1; i <= n; ++ i)
    {
        cin >> ri[i];
        if (i & 1) L[i] = change(L[i - 1], -(ri[i] - ri[i - 1]));
        else L[i] = change(L[i - 1], ri[i] - ri[i - 1]);
    }
    cin >> q;
    while (q --)
    {
        int t, a;
        cin >> t >> a;
        P3 nw;
        int p = upper_bound(ri, ri + n + 1, t) - ri - 1;
        if ((p + 1) & 1) nw = change(L[p], -(t - ri[p]));
        else nw = change(L[p], t - ri[p]);
        if (a <= nw.a) cout << nw.c;
        else if (a <= nw.a + nw.b) cout << nw.c + (a - nw.a);
        else cout << nw.c + nw.b;
        cout << endl;
    }
}