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;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
int x, k, r[MX], q, t, a;
struct Sand{int xl, xr, yl, yr;} sd[MX][18];
int calc(Sand& a, int x){return min(a.yr, a.yl + max(0,x-a.xl));}
int rcalc(Sand& a, int y){return min(a.xr, a.xl + max(0,y-a.yl));}
Sand merge(Sand& a, Sand& b){
	int ml = max(b.xl,a.yl), mr = min(b.xr,a.yr);
	return (Sand){rcalc(a,ml),rcalc(a,mr),calc(b,ml),calc(b,mr)};
}
int main(){
	scanf("%d%d", &x, &k);
	rep(i,k){
		scanf("%d", r+i+1);
		int dr = r[i+1] - r[i];
		if(i&1) sd[i][0] = (Sand){0,max(0,x-dr),min(x,dr),x};
		else sd[i][0] = (Sand){min(x,dr),x,0,max(0,x-dr)};
	}
	for(int j = 1; j < 18; j++) rep(i,k) if(i + (1<<j) <= k){
		sd[i][j] = merge(sd[i][j-1],sd[i+(1<<(j-1))][j-1]);
	}
	scanf("%d", &q);
	rep(i,q){
		scanf("%d%d", &t, &a);
		int it = (int)(upper_bound(r,r+k+1,t) - r)-1, c = 0;
		for(int j = 0; it >> j; j++) if(it>>j&1){
			a = calc(sd[c][j],a);
			c += 1 << j;
		}
		if(it&1) a = min(x,a+t-r[it]);
		else a = max(0,a-(t-r[it]));
		printf("%d\n", a);
	}
}