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;

struct Event {
    int t, d, a;
    Event() {}
    Event(int _t, int _d, int _a) : t(_t), d(_d), a(_a) {}
    bool operator<(const Event& _oth) const {
        return t < _oth.t;
    }
};

void solve() {
    int R, K;
    cin >> R >> K;
    vector<Event> es;
    for (int _ = 0; _ < K; _++) {
        int t; cin >> t;
        es.emplace_back(t, -1, 0);
    }
    int Q; cin >> Q;
    for (int _ = 0; _ < Q; _++) {
        int t,a;
        cin >> t >> a;
        es.emplace_back(t, 1, a);
    }
    sort(es.begin(), es.end());
    
    int l = 0, r = R, x = 0;
    int t = 0, d = -1;
    for (auto& e: es) {
        x += d * (e.t - t); t = e.t; d *= e.d;
        l = min(max(l,-x), -x+R);
        r = min(max(r,-x), -x+R);

        if (e.d > 0) {
            int a = min(max(e.a, l), r);
            cout << (a + x) << "\n";
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}
