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 <cstdio>
#include <cctype>
#include <algorithm>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

int n,m,a[100100],l[100100],r[100100],w[100100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()));
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

int main()
{
    scanf("%d%d",&m,&n),r[0]=m;
    repu(i,1,n)
    {
        int t=(a[i]=getint())-a[i-1];
        if (!(i&1))
            if (l[i-1]+w[i-1]+t>m)
                l[i]=r[i]=m,w[i]=0;
            else
                l[i]=l[i-1],r[i]=min(r[i-1],m-w[i-1]-t),w[i]=w[i-1]+t;
        else
            if (r[i-1]+w[i-1]-t>=0)
                l[i]=max(l[i-1],t-w[i-1]),r[i]=r[i-1],w[i]=w[i-1]-t;
    }
    for (int q=getint(),i=0; q--;)
    {
        int t=getint(),x=getint();
        for (; i<n && a[i+1]<=t; ++i);
        int ans=max(min(x+w[i],r[i]+w[i]),l[i]+w[i]);
        if (i&1)
            ans=min(m,ans+t-a[i]);
        else
            ans=max(0,ans+a[i]-t);
        printf("%d\n",ans);
    }
    return 0;
}