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>
#define DIM 100005
using namespace std;
int x, n, q, t, val, sol, u, i, sum;
int v[DIM], a[DIM], amax[DIM], vmin[DIM];
int main(){
    cin>> x >> n;
    for(i = 1; i <= n; i++){
        cin>> v[i];
    }
    amax[0] = x;
    for(i = 1; i <= n; i++){
        if(i % 2 == 1){
            a[i] = max(0, a[i - 1] - (v[i] - v[i - 1]) );
            amax[i] = max(0, amax[i - 1] - (v[i] - v[i - 1]) );
            sum -= v[i] - v[i - 1];
        }
        else{
            a[i] = min(x, a[i - 1] + v[i] - v[i - 1]);
            amax[i] = min(x, amax[i - 1] + v[i] - v[i - 1]);
            sum += v[i] - v[i - 1];
        }
        vmin[i] = min(vmin[i - 1], sum);
    }
    v[n + 1] = 1000000001;
    u = 0;
    cin>> q;
    for(; q; q--){
        cin>> t >> val;
        while(v[u + 1] <= t){
            u++;
        }
        sol = min(a[u] + max(0, val + vmin[u]), amax[u]);
        if(u % 2 == 1){
            sol = min(x, sol + t - v[u]);
        }
        else{
            sol = max(0, sol - (t - v[u]) );
        }
        cout<< sol <<"\n";
    }
    return 0;
}
