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>
#define ll long long
using namespace std;
ll x;
ll r[100005]={0};
ll fun(ll t,ll low,ll high)
{
    if(t<low)return low;
    if(t>high)return high;
    return t; 
}
int main()
{
    int q,k;
    scanf("%lld%d",&x,&k);
    for(int i=1;i<=k;i++)
    scanf("%lld",&r[i]);
    scanf("%d",&q);
    ll l=0,h=x;
    ll add=0;
    int now=1;
    int flag=-1;
    while(q--)
    {
        ll tt,a;
        scanf("%lld%lld",&tt,&a);
        while(tt>=r[now]&&now<=k)
        {
            ll xx=flag*(r[now]-r[now-1]);
            add=add+xx;
            h=fun(h+xx,0,x);
            l=fun(l+xx,0,x);
            flag=-1*flag;
            now++;
        }
        ll ans=fun(add+a,l,h);
        ans=fun(ans+flag*(tt-r[now-1]),0,x);
        printf("%lld\n",ans);
    }
    return 0;
}