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;
const int maxn = 1e5+5;
int X, K, Q, r[maxn], t[maxn], a[maxn], ub, lb, b, t1, t2, add, tmp, diff, sgn;
void Add(int &n, int num){
    n+=num;
    n = max(0, min(X, n));
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> X >> K;
    for(int i = 1; i <= K; i++) cin >> r[i];
    cin >> Q;
    for(int i = 1; i <= Q; i++) cin >> t[i] >> a[i];
    t1 = 0, t2 = 1;
    lb = b = 0, ub = X, sgn = -1;
    while(t2 <= Q){
        if(t[t2] > r[t1+1] && t1 < K){
            t1++;
            add = sgn*(r[t1] - r[t1-1]);
            Add(lb, add); Add(ub, add);
            b+=add;
            sgn*=-1;
        }
        else{
            diff = (t[t2] - r[t1]);
            tmp = max(lb, min(ub, a[t2]+b));
            Add(tmp, diff*sgn);
            cout << tmp << "\n";
            t2++;
        }
    }
    return 0;
}