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;

int main() {
    std::ios::sync_with_stdio(false);
    int x;cin >> x;
    int k;cin >> k;
    vector<int> r, a, b, c;
    r.push_back(0);
    a.push_back(0);
    b.push_back(x);
    c.push_back(0);
    for(int i=0;i<k;i++){
        int rr;cin >> rr;
        int d = rr-r[i];
        int aa, bb, cc;
        if(i%2 == 0){
            aa = min(b[i], a[i] + max(0, 0-(c[i]-d)));
            bb = b[i];
            cc = max(0, c[i]-d);
        }else{
            aa = a[i];
            bb = max(a[i], b[i] - max(0, c[i]+(b[i]-a[i])+d-x));
            cc = min(x, c[i]+d);
        }
        r.push_back(rr);
        a.push_back(aa);
        b.push_back(bb);
        c.push_back(cc);
    }
    int q;cin >> q;
    for(int i=0;i<q;i++){
        int t, aa;cin >> t >> aa;
        int j = int(upper_bound(r.begin(), r.end(), t) - r.begin()) - 1;
        int d = t-r[j];
        int f;
        if(aa <= a[j]){
            f = c[j];
        }else if(aa <= b[j]){
            f = c[j] + (aa - a[j]);
        }else{
            f = c[j] + (b[j]-a[j]);
        }
        if(j%2 == 0){
            cout << max(0, f-d) << endl;
        }else{
            cout << min(x, f+d) << endl;
        }
    }
    return 0;
}
