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>

using namespace std;

int main() {
    const int z = 100001;
    int x, k, r[z], s[z], o[z], w[z], q, i;
    cin >> x;
    cin >> k;
    int v = -1;
    r[0] = w[0] = o[0] = 0;
    s[0] = x;
    for (i = 1; i <= k; i++, v *= -1) {
        cin >> r[i];
        int d = (r[i] - r[i-1]) * v;
        w[i] = min(max(w[i-1] + d, 0), x);
        s[i] = max(min(s[i-1] + d, x), 0);
        o[i] = min(o[i-1] - min(0, w[i-1] + d), x);
    }
    cin >> q;
    i = 0;
    while(q--) {
        int t, a;
        cin >> t >> a;
        if (t >= r[k]) i = k;
        else {
            int n = k;
            while (i < n - 1) {
                int c = i + (n - i) / 2;
                if (r[c] < t) i = c;
                else n = c;
            }
        }
        int ans = min(w[i] + max(a - o[i], 0), s[i]) + (t - r[i]) * ((i % 2) * 2 - 1);
        cout << min(max(ans, 0), x) << endl;
    }
}
