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 main(void)
{
    int x,k,q;
    vector<int> r;
    vector<int> a,t;

    cin>>x>>k;
    
    for(int i = 0;i < k;i++)
    {
	int temp;
	cin>>temp;
	r.push_back(temp);
    }
    cin>>q;

    for(int i = 0;i < q;i++)
    {
	int ta,tt;

	cin>>tt>>ta;
	t.push_back(tt);
	a.push_back(ta);
    }

    int target = 0;
    int position = 0;
    int ctime = 0;
    int sign = -1;
    int dx = 0;
    int minimum = 0,maximum = x;

    while(position < q)
    {
	if(target == k || t[position] < r[target])
	{
	    int f = max(minimum,min(maximum,a[position]+dx)) + sign*(t[position] - ctime);
	    f = max(0,min(x,f));
	    cout<<f<<endl;
	    position++;
	} else {
	    //Calc variaion
	   int diff = sign*(r[target] - ctime);

	   minimum += diff;
	   maximum += diff;
	   dx += diff;

	   minimum = min(x,minimum);
	   if(minimum < 0)
	       minimum = 0;
	   maximum = min(x,maximum);
	   if(maximum < 0)
	       maximum = 0;

	   sign *= -1;
	   ctime = r[target];
	   target++;
	}

    }
    return 0;
}

