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>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+50;
typedef long long ll;
int r[maxn];
int x,k,q,t,a;
int sgn=-1;
int pos=1,und,ds;
ll mx,mn,all;
void cal(ll &y)
{
    if(y<0)
        y=0;
    if(y>x)
        y=x;
}
void solve()
{
    while(r[pos]<=t&&pos<=k)
    {
        ll ds=(r[pos]-r[pos-1])*sgn;
        mx+=ds;
        mn+=ds;
        all+=ds;
        cal(mx);
        cal(mn);
        pos++;
        sgn*=-1;
        //cout << all << endl;
    }
    ll ans=a+all;
    if(ans>mx)
        ans=mx;
    if(ans<mn)
        ans=mn;
    ll temp=(t-r[pos-1])*sgn;
    ans+=temp;
    cal(ans);
    printf("%lld\n",ans);
}
int main()
{
    //ios::sync_with_stdio(false);
    scanf("%d%d",&x,&k);
    for(int i=1; i<=k; i++)
    {
        scanf("%d",&r[i]);
        //cin>>r[i];
    }
    pos=1;
    mx=x,mn=0,all=0;
    sgn=-1;
    r[0]=0;
    scanf("%d",&q);
    while(q--)
    {

        //cin>>t>>a;
        scanf("%d%d",&t,&a);
        solve();
    }
    //cout << "Hello world!" << endl;
    return 0;
}
