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;
using lint = long long;
using pi = pair<int, int>;
const int MAXN = 1000005;

int x, n;

int t[MAXN], p[MAXN], q[MAXN], l[MAXN];
int main(){
	scanf("%d %d",&x,&n);
	l[0] = x;
	for(int i=1; i<=n; i++){
		scanf("%d",&t[i]);
		int del = t[i] - t[i-1];
		tie(p[i], q[i], l[i]) = make_tuple(p[i-1], q[i-1], l[i-1]);
		if(i & 1){
			int dx = min(del, q[i]);
			q[i] -= dx;
			del -= dx;
			dx = min(del, l[i]);
			p[i] += dx;
			l[i] -= dx;
		}
		else{
			int dx = min(del, x - q[i] - l[i]);
			q[i] += dx;
			del -= dx;
			dx = min(del, l[i]);
			q[i] += dx;
			l[i] -= dx;
		}
	//	printf("%d %d %d\n", p[i], q[i], l[i]);
	}
	int qr;
	scanf("%d",&qr);
	for(int i=0; i<qr; i++){
		int ti, v; scanf("%d %d",&ti,&v);
		int pos= upper_bound(t, t + n + 1, ti) - t - 1;
		int now = 0;
		if(v <= p[pos]) now = q[pos];
		else now = q[pos] + min(l[pos], v - p[pos]);
		ti -= t[pos];
		if(pos % 2 == 0) now = max(0, now - ti);
		else now = min(x, now + ti);
		printf("%d\n", now);
	}

}

