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;
typedef long long LL;

int main(){
	LL x, k;
	cin >> x >> k;
	LL r[k];
	for(int i = 0; i < k; i++) cin >> r[i];
	LL q;
	cin >> q;
	vector<LL> t;
	vector<LL> a;
	for(int i = 0; i < q; i++){
		LL p, r;
		cin >> p >> r;
		t.push_back(p);
		a.push_back(r);
	}

	LL minv = 0;
	LL maxv = x;
	LL dx = 0;
	LL sgn = -1;

	LL s1 = 0;
	LL s2 = 0;
	LL lastt = 0;
	while(s2 < t.size()){
		if(s1 == k || t[s2] < r[s1]){
			// answer
			LL f = max(minv, min(maxv, a[s2] + dx));
			f += sgn*(t[s2] - lastt);
			f = max(0LL, min(x, f));
			cout << f << endl;
			s2++;
		} else {
			LL diff = r[s1] - lastt;
			diff *= sgn;
			minv += diff;
			maxv += diff;
			dx += diff;
			minv = max(0LL, min(x, minv));
			maxv = max(0LL, min(x, maxv));
			sgn *= -1;
			lastt = r[s1];
			s1++;
			//cout << minv << " " << maxv << " " << dx << " " << sgn << endl;
		}
	}
}
