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 X,n,q;
int r[100001],t[100000],a[100000];

int main() {
    scanf("%d%d",&X,&n);
    for(int i=1;i<=n;i++) scanf("%d",r+i);
    scanf("%d",&q);
    for(int i=0;i<q;i++) scanf("%d%d",t+i,a+i);

    int sl=0,sr=X,ll=0,rr=X;
    for(int i=0,j=1;i<q;i++) {
        for(;j<=n && r[j]<=t[i];j++) {
            int d=r[j]-r[j-1];
            if(j%2==1) {
                sl += max(d-ll,0);
                sl = min(sl,sr);
                ll = max(ll-d,0);
                rr = max(rr-d,0);
            } else {
                sr -= max(d-(X-rr), 0);
                sr = max(sl,sr);
                rr = min(rr+d,X);
                ll = min(ll+d,X);
            }
        }
        a[i] = max(a[i],sl);
        a[i] = min(a[i],sr);

        int v=ll+(a[i]-sl);
        if(j%2==1) v-=t[i]-r[j-1];
        else v+=t[i]-r[j-1];
        v = min(v,X); v=max(v,0);
        printf("%d\n",v);
    }
    
    return 0;
}
