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<stdio.h>
#include<queue>
#include<math.h>
#include<time.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#define LL long long
#define mem(a,b) memset(a,b,sizeof(a))
#define lowbit(a) a&(-a)
#define PI acos(-1)
#define shortime(a)  std::ios::sync_with_stdio(a);
using  namespace std;
const LL inf=16777216;
//long long cmp(node a,node b){ if(a.x==b.x) return a.r>b.r;return a.x>b.x;}
int maxn (int a,int b,int c){return max(max(a,b),max(b,c));}
int judge(int x,int y,int z)
{
    if(x<y) return y;
    if(x>z) return z;
    return x;
}
int main()
{
    int x,k,a[100005];
    scanf("%d%d",&x,&k);
    for(int i=1;i<=k;i++)
    {
        scanf("%d",&a[i]);
    }
    int n,l=0,r=x,num=1;
    scanf("%d",&n);
    int flag=-1,sum=0;
    for(int i=0;i<n;i++)
    {
        int lr,t,a1;
        scanf("%d%d",&t,&a1);
        while(t>=a[num]&&num<=k)
        {
            lr=flag*(a[num]-a[num-1]);
            sum+=lr;
            l=judge(lr+l,0,x);
            r=judge(lr+r,0,x);
            num++;
            flag=-flag;
        }
        printf("%d\n",judge(flag*(t-a[num-1])+judge(a1+sum,l,r),0,x));
    }
    return 0;
}
