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 <algorithm>
#include <iostream>
#include <map>
#include <vector>
using namespace std;

int main()
{
    cin.tie(0);

    int x, k;
    cin >> x >> k;

    vector<int> rs(k);
    for (auto& r : rs)
        cin >> r;
    rs.push_back(1e9 + 1);

    int qn;
    cin >> qn;
    vector<pair<int, int>> qs(qn);
    for (auto& q : qs) {
        cin >> q.first >> q.second;
    }

    int lo = 0, hi = x;
    int qi = 0;
    int prev = 0;
    int base = 0;

    for (int i = 0; i < rs.size(); i++) {
        // cerr << "hi: " << hi << ", lo: " << lo << ", base: " << base << endl;

        int dir = i % 2 == 0 ? -1 : 1;
        int r = rs[i];
        while (qi < qs.size() && qs[qi].first < r) {
            int t = qs[qi].first - prev;
            int b = min(hi - lo, max(0, qs[qi].second - base)) + lo;
            int ans = min(x, max(0, b + t * dir));
            cout << ans << endl;
            qi++;
        }

        int ma = min(x, max(0, hi + (r - prev) * dir));
        int mi = min(x, max(0, lo + (r - prev) * dir));

        base += (mi - (r - prev) * dir) - lo;

        hi = ma;
        lo = mi;

        prev = r;
    }

    return 0;
}
