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=100010;
int n,m,a[maxn];
int calc(int l,int r,int x)
{
	return max(l,min(r,x));
}
int main()
{
    int X;
    scanf("%d%d",&X,&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    int L,R=X;
    int t=0,k=0,s=-1,x=0;
    int time,A;
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
	{
        scanf("%d%d",&time,&A);
        while(k<n&&a[k+1]<=time)
		{
            int dif=s*(a[k+1]-t);
            L=calc(0,X,L+dif);
            R=calc(0,X,R+dif);
            s*=-1;
            x+=dif;
            t=a[k+1];
            k++;
        }
        int T=time-t;
        A=calc(L,R,A+x);
        A=calc(0,X,A+s*T);
        printf("%d\n",A);
    }
    return 0;
}