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<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll>pii;
int main()
{
	ll gen;
	int num;
	scanf("%lld%d", &gen, &num);
	vector<pii>qs;
	for (int i = 0; i < num; i++)
	{
		ll z;
		scanf("%lld", &z);
		qs.push_back(make_pair(z, -1));
	}
	int query;
	scanf("%d", &query);
	for (int i = 0; i < query; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		qs.push_back(make_pair(za, zb));
	}
	sort(qs.begin(), qs.end());
	ll mina = 0;
	ll lb = 0, ub = gen;
	ll bef = 0;
	int d = 0;
	for (int i = 0; i < qs.size(); i++)
	{
		if (qs[i].second == -1)
		{
			ll t = qs[i].first - bef;
			if (d == 0)
			{
				if (t < lb)lb -= t, ub -= t;
				else if (t < ub)mina += t - lb, lb = 0, ub -= t;
				else lb = ub = 0;
			}
			else
			{
				if (t < gen - ub)lb += t, ub += t;
				else if (t < gen - lb)lb += t, ub = gen;
				else lb = ub = gen;
			}
			d ^= 1;
			bef = qs[i].first;
		}
		else
		{
			ll t = qs[i].first - bef;
			ll z;
			if (qs[i].second < mina)z = lb;
			else if (qs[i].second <= mina + ub - lb)z = lb + (qs[i].second - mina);
			else z = ub;
			if (d == 0)z = max(z - t, 0LL);
			else z = min(z + t, gen);
			printf("%lld\n", z);
		}
	}
}